Question 15 Marks
Solve the following equation for x:
$\tan^{-1}\frac{1}{4}+2\tan^{-1}\frac{1}{5}+\tan^{-1}\frac{1}{6}+\tan^{-1}\frac{1}{\text{x}}=\frac{\pi}{4}$
View full question & answer→Question 25 Marks
If $\sin^{-1}\text{x}+\sin^{-1}\text{y}=\frac{\pi}{3}$ and $\cos^{-1}\text{x}-\cos^{-1}\text{y}=\frac{\pi}{6},$ find the values of x and y.
Answer$\cos^{-1}\text{x}-\cos^{-1}\text{y}=\frac{\pi}{6}$
$\Rightarrow\frac{\pi}{2}-\sin^{-1}\text{x}-\frac{\pi}{2}+\sin^{-1}\text{y}=\frac{\pi}{6}$
$\Big[\because\ \cos^{-1}\text{x}=\frac{\pi}{2}-\sin^{-1}\text{x}\Big]$
$\Rightarrow-\big(\sin^{-1}\text{x}-\sin^{-1}\text{y}\big)=\frac{\pi}{6}$
$\Rightarrow\sin^{-1}\text{x}-\sin^{-1}\text{y}=-\frac{\pi}{6}$
Solving $\sin^{-1}\text{x}+\sin^{-1}\text{y}=\frac{\pi}{3}$ and $\sin^{-1}\text{x}-\sin^{-1}\text{y}=-\frac{\pi}{6},$ we will get
$2\sin^{-1}\text{x}=\frac{\pi}{6}$
$\Rightarrow\sin^{-1}\text{x}=\frac{\pi}{12}$
$\Rightarrow\text{x}=\sin\frac{\pi}{12}=\frac{\sqrt3-1}{2\sqrt2}$
and
$\sin^{-1}\text{y}=\frac{\pi}{3}-\sin^{-1}\text{x}$
$\Rightarrow\sin^{-1}\text{y}=\frac{\pi}{3}-\frac{\pi}{12}$
$\Rightarrow\sin^{-1}\text{y}=\frac{\pi}{4}$
$\Rightarrow\text{y}=\sin\frac{\pi}{4}=\frac{1}{\sqrt2}$
View full question & answer→Question 35 Marks
Prove that $2\tan^{-1}\bigg(\sqrt{\frac{\text{a}-\text{b}}{\text{a}+\text{b}}}\tan\frac{\theta}{2}\bigg)=\cos^{-1}\Big(\frac{\text{a}\cos\theta+b}{\text{a}+\text{b}\cos\theta}\Big)$
Answer$\text{L.H.S}=2\tan^{-1}\bigg(\sqrt{\frac{\text{a}-\text{b}}{\text{a}+\text{b}}}\tan\frac{\theta}{2}\bigg)$
$=\cos^{-1}\begin{Bmatrix}\frac{1-\Big(\sqrt{\frac{\text{a}-\text{b}}{\text{a}+\text{b}}}\tan\frac{\theta}{2}\Big)^2}{1+\Big(\sqrt{\frac{\text{a}-\text{b}}{\text{a}+\text{b}}}\tan\frac{\theta}{2}\Big)^2}\end{Bmatrix}$ $\Big[\because\ 2\tan^{-1}(\text{x})=\cos^{-1}\Big\{\frac{1-\text{x}^2}{1-\text{x}^2}\Big\}\Big]$
$=\cos^{-1}\Bigg\{\frac{1-\frac{\text{a}-\text{b}}{\text{a}++\text{b}}\tan^{2}\frac{\theta}{2}}{1+\frac{\text{a}-\text{b}}{\text{a}++\text{b}}\tan^{2}\frac{\theta}{2}}\Bigg\}$
$=\cos^{-1}\Bigg\{\frac{\text{a}+\text{b}-(\text{a}-\text{b})\tan^{2}\frac{\theta}{2}}{\text{a}+\text{b}+(\text{a}-\text{b})\tan^{2}\frac{\theta}{2}}\Bigg\}$
$=\cos^{-1}\Bigg\{\frac{\text{a}+\text{b}-\text{a}\tan^2\frac{\theta}{2}+\text{b}\tan^2\frac{\theta}{2}}{\text{a}+\text{b}+\text{a}\tan^2\frac{\theta}{2}-\text{b}\tan^2\frac{\theta}{2}}\Bigg\}$
$=\cos\begin{Bmatrix}\frac{\text{a}\Big(1-\tan^2\frac{\theta}{2}\Big)+\text{b}\Big(1+\tan^2\frac{\theta}{2}\Big)}{\text{a}\Big(1+\tan^2\frac{\theta}{2}\Big)+\text{b}\Big(1-\tan^2\frac{\theta}{2}\Big)}\end{Bmatrix}$
$=\cos^{-1}\begin{Bmatrix}\frac{\text{a}\Bigg(\frac{1-\tan^2\frac{\theta}{2}}{1+\tan^2\frac{\theta}{2}}\Bigg)+\text{b}\Bigg(\frac{1+\tan^2\frac{\theta}{2}}{1+\tan^2\frac{\theta}{2}}\Bigg)}{\text{a}\Bigg(\frac{1+\tan^2\frac{\theta}{2}}{1+\tan^2\frac{\theta}{2}}\Bigg)+\text{b}\Bigg(\frac{1-\tan^2\frac{\theta}{2}}{1-\tan^2\frac{\theta}{2}}\Bigg)}\end{Bmatrix}$ $\Big[\text{Dividing N' and D' by }1+\tan^2\frac{\theta}{2}\Big]$
$=\cos^{-1}\begin{Bmatrix}\frac{\text{a}\Bigg(\frac{1-\tan^2\frac{\theta}{2}}{1+\tan^2\frac{\theta}{2}}\Bigg)+\text{b}}{\text{a}+\text{b}\Bigg(\frac{1-\tan^2\frac{\theta}{2}}{1-\tan^2\frac{\theta}{2}}\Bigg)}\end{Bmatrix}$
$=\cos^{-1}\Big(\frac{\text{a}\cos\theta+b}{\text{a}+\text{b}\cos\theta}\Big)=\text{R.H.S}$
View full question & answer→Question 45 Marks
For any a, b, x, y > 0, prove that:
$\frac{2}{3}\tan^{-1}\Big(\frac{3\text{a}\text{b}^2-\text{a}^3}{\text{b}^3-3\text{a}^2\text{b}}\Big)+\frac{2}{3}\tan^{-1}\Big(\frac{3\text{x}\text{y}^2-\text{x}^3}{\text{y}^3-3\text{x}^2\text{y}}\Big)=\tan^{-1}\frac{2\alpha\beta}{\alpha^2-\beta^2}$
where $\alpha=-\text{ax}+\text{by},\beta=\text{bx}+\text{ay}$
AnswerLet $\text{a}=\text{b}\tan\text{m}$ and $\text{x}=\text{y}\tan\text{n}$
Then
$\frac{2}{3}\tan^{-1}\Big(\frac{3\text{a}\text{b}^2-\text{a}^3}{\text{b}^3-3\text{a}^2\text{b}}\Big)+\frac{2}{3}\tan^{-1}\Big(\frac{3\text{x}\text{y}^2-\text{x}^3}{\text{y}^3-3\text{x}^2\text{y}}\Big)$
$=\frac{2}{3}\tan^{-1}\Big(\frac{3\text{b}^3\tan\text{m}-\text{b}^3\tan^3\text{m}}{\text{b}^3-3\text{b}^3\tan^2\text{m}}\Big)+\frac{2}{3}\tan^{-1}\Big(\frac{3\text{y}^3\tan\text{n}-\text{y}^3\tan^3\text{n}}{\text{y}^3-3\text{y}^3\tan^2\text{n}}\Big)$
$=\frac{2}{3}\tan^{-1}\Big(\frac{3\tan\text{m}-\tan^3\text{m}}{1-3\tan^2\text{m}}\Big)+\frac{2}{3}\tan^{-1}\Big(\frac{3\tan\text{n}-\tan^3\text{n}}{1-\tan^2\text{n}}\Big)$
$=\frac{2}{3}\tan^{-1}(\tan3\text{m})+\frac{2}{3}\tan^{-1}(\tan3\text{n})$ $[\because\ \text{a}=\text{b}\tan\text{m},\text{x}=\text{y}=\tan\text{n}]$
$=2\tan^{-1}\Bigg(\frac{\frac{\text{a}}{\text{b}}+\frac{\text{x}}{\text{y}}}{1-\frac{\text{a}}{\text{b}}\frac{\text{x}}{\text{y}}}\Bigg)$
$=2\tan^{-1}\Big(\frac{\text{ay}+\text{bx}}{\text{by}-\text{ax}}\Big)$
$=\tan^{-1}\begin{Bmatrix}\frac{2\frac{\text{ay}+\text{bx}}{\text{by}-\text{ax}}}{1-\Big(\frac{\text{ay}+\text{bx}}{\text{by}-\text{ax}}\Big)^2}\end{Bmatrix}$
$=\tan^{-1}\Bigg\{\frac{2(\text{ay}+\text{bx})(\text{by}-\text{ax})}{(\text{by}-\text{ax})^2-(\text{ay}+\text{bx})^2}\Bigg\}$
$=\tan^{-1}\Big\{\frac{2\alpha\beta}{\alpha^2-\beta^2}\Big\}$ $[\because\ \beta=\text{bx}+\text{ay},\alpha=-\text{ax}+\text{by}]$
View full question & answer→Question 55 Marks
Prove that: $\tan^{-1}\frac{2\text{a}\text{b}}{\text{a}^2-\text{b}^2}+\tan^{-1}\frac{2\text{xy}}{\text{x}^2-\text{y}^2}=\tan^{-1}\frac{2\alpha\beta}{\alpha^2-\beta^2},$where $\alpha=\text{ax}-\text{by}$ and $\beta=\text{ay}+\text{bx}.$
Answer$\tan^{-1}\frac{2\text{a}\text{b}}{\text{a}^2-\text{b}^2}+\tan^{-1}\frac{2\text{xy}}{\text{x}^2-\text{y}^2}=\tan^{-1}\frac{2\alpha\beta}{\alpha^2-\beta^2}$ as $\alpha=\text{ax}-\text{by},\beta=\text{ay}+\text{bx}$
$\text{L.H.S}=\tan^{-1}\frac{2\text{a}\text{b}}{\text{a}^2-\text{b}^2}+\tan^{-1}\frac{2\text{xy}}{\text{x}^2-\text{y}^2}$
$=\tan^{-1}\begin{bmatrix}\frac{\frac{2\text{a}\text{b}}{\text{a}^2-\text{b}^2}+\frac{2\text{xy}}{\text{x}^2-\text{y}^2}}{1-\Big(\frac{2\text{ab}}{\text{a}^2-\text{b}^2}\Big)\Big(\frac{2\text{xy}}{\text{x}^2-\text{y}^2}\Big)}\end{bmatrix}$ $\Big\{\text{Since},\tan^{-1}\text{x}+\tan^{-1}\text{y}=\tan^{-1}\frac{\text{x}+\text{y}}{1-\text{xy}}\Big\}$
$=\tan^{-1}\begin{bmatrix}\frac{\frac{2\text{a}\text{b}\text{x}^2-2\text{a}\text{b}\text{y}^2+2\text{xy}\text{a}^2-2\text{xyb}^2}{\big(\text{a}^2-\text{b}^2\big)\big(\text{x}^2-\text{y}^2\big)}}{\frac{\text{a}^2\text{x}^2-\text{a}^2\text{y}^2-\text{b}^2\text{y}^2+\text{b}^2\text{y}^2-4\text{abxy}}{\big(\text{a}^2-\text{b}^2\big)\big(\text{x}^2-\text{y}^2\big)}}\end{bmatrix}$
$=\tan^{-1}\Bigg[\frac{2\big(\text{ab}\text{x}^2+\text{xy}\text{a}^2-\text{ab}\text{y}^2-\text{xy}\text{b}^2\big)}{\text{a}^2\text{x}^2+\text{b}^2\text{y}^2-2\text{abxy}-\text{a}^2\text{y}^2-\text{b}^2\text{y}^2-2\text{abxy}} \Bigg]$
$=\tan^{-1}\Bigg[\frac{2\{\text{ax}(\text{bx}+\text{ay})-\text{by}(\text{ay}+\text{bx})\}}{(\text{ax}-\text{by})^2-\big(\text{a}^2\text{y}^2+\text{b}^2\text{x}^2+2\text{abxy}\big)}\Bigg]$
$=\tan^{-1}\Bigg[\frac{2(\text{bx}+\text{ay})(\text{ax}-\text{by})\}}{(\text{ax}-\text{by})^2-(\text{bx}+\text{ay})^2}\Bigg]$
$=\tan^{-1}\Big[\frac{2\alpha\beta}{\alpha^2-\beta^2}\Big]$ $\{\text{Since},\alpha=\text{ax}-\text{by},\beta=\text{ay}+\text{bx}\}$
Hence,
$\tan^{-1}\bigg(\frac{2\text{a}\text{b}}{\text{a}^2-\text{b}^2}\bigg)+\tan^{-1}\bigg(\frac{2\text{xy}}{\text{x}^2-\text{y}^2}\bigg)=\tan^{-1}\bigg(\frac{2\alpha\beta}{\alpha^2-\beta^2}\bigg)$
View full question & answer→Question 65 Marks
Prove the following results:
$\sin^{-1}\frac{5}{13}+\cos^{-1}\frac{3}{5}=\tan^{-1}\frac{63}{16}$
Answer$\text{L.H.S }\sin^{-1}\frac{5}{13}+\cos^{-1}\frac{3}{5}$
$=\sin^{-1}\frac{5}{13}+\cos^{-1}\frac{3}{5}$
$=\sin^{-1}\frac{5}{13}+\sin^{-1}\sqrt{1-\Big(\frac{3}{5}\Big)^2}$ $\Big[\because\ \sin^{-1}\text{x}=\cos^{-1}\sqrt{1-\text{x}^2}\Big]$
$=\sin^{-1}\frac{5}{13}+\sin^{-1}\frac{4}{5}$
$=\sin^{-1}\Bigg[\frac{5}{13}\sqrt{1-\Big(\frac{4}{5}\Big)^2}+\frac{4}{5}\sqrt{1-\Big(\frac{5}{13}\Big)^2}\Bigg]$
$\Big[\because\ \sin^{-1}\text{x}+\sin^{-1}\text{y}=\sin^{-1}\Big(\text{x}\sqrt{1-\text{y}^2}+\text{y}\sqrt{1-\text{x}^2}\Big)\Big]$
$=\sin^{-1}\Big(\frac{5}{13}\times\frac{3}{5}+\frac{4}{5}\times\frac{12}{13}\Big)$
$=\sin^{-1}\Big(\frac{3}{13}+\frac{48}{65}\Big)$
$=\sin^{-1}\Big(\frac{63}{65}\Big)$
$=\tan^{-1}\begin{pmatrix}\frac{\frac{63}{65}}{\sqrt{1-\frac{63^2}{65}}}\end{pmatrix}$ $\Big[\because\ \sin^{-1}\text{x}=\tan^{-1}\Big(\frac{\text{x}}{\sqrt{1-\text{x}^2}}\Big)\Big]$
$=\tan^{-1}\Bigg(\frac{\frac{63}{65}}{\frac{16}{65}}\Bigg)$
$=\tan^{-1}\Big(\frac{63}{16}\Big)=\text{R.H.S}$
View full question & answer→Question 75 Marks
Show that $2\tan^{-1}\text{x}+\sin^{-1}\frac{2\text{x}}{1+\text{x}^2}$ is constant for $\text{x}\geq1,$ find that constant.
Answer$2\tan^{-1}\text{x}+\sin^{-1}\Big(\frac{2\text{x}}{1+\text{x}^2}\Big)$
- For x > 1
$ =2\tan^{-1}\text{x}+\sin^{-1}\Big(\frac{2\text{x}}{1+\text{x}^2}\Big)$
$=\pi-\sin^{-1}\Big(\frac{2\text{x}}{1+\text{x}^2}\Big)+\sin^{-1}\Big(\frac{2\text{x}}{1+\text{x}^2}\Big)$ $\Big[\because\ 2\tan^{-1}\text{x}=\pi-\sin^{-1}\Big(\frac{2\text{x}}{1+\text{x}^2}\Big),\text{x}>1\Big]$
$=\pi$
- For x = 1
$=2\tan^{-1}\text{x}+\sin^{-1}\Big(\frac{2\text{x}}{1+\text{x}^2}\Big)$
$=2\tan^{-1}(1)+\sin^{-1}\bigg(\frac{2(1)}{1+(1)^2}\bigg)$
$=2\tan^{-1}(1)+\sin^{-1}(1)$
$=2\Big(\frac{\pi}{4}\Big)+\frac{\pi}{2}$
$=\pi$ View full question & answer→Question 85 Marks
Prove the following results:
$2\sin^{-1}\frac{3}{5}-\tan^{-1}\frac{17}{31}=\frac{\pi}{4}$
Answer$\text{L.H.S}=2\sin^{-1}\Big(\frac{3}{5}\Big)-\tan^{-1}\Big(\frac{17}{31}\Big)$
$=2\tan^{-1}\Bigg(\frac{\frac{3}{4}}{\sqrt{1-\frac{9}{25}}}\Bigg)-\tan^{-1}\Big(\frac{17}{31}\Big)$ $\Big[\because\ \sin^{-1}\text{x}=\tan^{-1}\frac{\text{x}}{\sqrt{1-\text{x}^2}}\Big]$
$=2\tan^{-1}\Bigg(\frac{\frac{3}{5}}{\frac{4}{5}}\Bigg)-\tan^{-1}\Big(\frac{17}{31}\Big)$
$=2\tan^{-1}\Big(\frac{3}{4}\Big)-\tan^{-1}\Big(\frac{17}{31}\Big)$
$=\tan^{-1}\Bigg\{\frac{2\times\frac{3}{4}}{1-\big(\frac{3}{4}\big)^2}\Bigg\}-\tan^{-1}\frac{17}{31}$ $\Big[\because\ 2\tan^{-1}\text{x}=\tan^{-1}\Big\{\frac{2\text{x}}{1-\text{x}^2}\Big\}\Big]$
$=\tan^{-1}\Bigg\{\frac{\frac{3}{2}}{\frac{7}{16}}\Bigg\}-\tan^{-1}\frac{17}{31}$
$=\sin^{-1}\Big(\frac{24}{7}\Big)+\tan^{-1}\Big(\frac{17}{31}\Big)$
$=\tan^{-1}\Bigg(\frac{\frac{24}{7}-\frac{17}{31}}{1+\frac{24}{7}\times\frac{17}{31}}\Bigg)$
$\Big[\because\ \tan^{-1}\text{x}-\tan^{-1}\text{y}=\tan^{-1}\Big(\frac{\text{x}-\text{y}}{1+\text{xy}}\Big)\Big]$
$=\tan^{-1}\Bigg(\frac{\frac{625}{217}}{\frac{625}{217}}\Bigg)$
$=\tan^{-1}(1)=\frac{\pi}{4}=\text{R.H.S}$
View full question & answer→Question 95 Marks
Solve the following equation for x:
$\cot^{-1}\text{x}-\cot^{-1}(\text{x}+2)=\frac{\pi}{12},\text{x}>0$
Answer$\Rightarrow\cot^{-1}(\text{x})-\cot^{-1}(\text{x}+2)=\frac{\pi}{12}$
$\Rightarrow \tan^{-1}\Big(\frac{1}{\text{x}}\Big)+\cot^{-1}\Big(\frac{1}{\text{x}+2}\Big)=\frac{\pi}{2}$
$\Big[\because\ \cot^{-1}\text{x}=\tan^{-1}\frac{1}{\text{x}}\Big]$
$\Rightarrow\tan^{-1}\Bigg(\frac{\frac{1}{\text{x}}-\frac{1}{\text{x}+2}}{1+\frac{1}{\text{x}(\text{x}+2)}}\Bigg)=\frac{\pi}{12}$
$\Rightarrow\Bigg(\frac{\frac{2}{\text{x}(\text{x}+2)}}{\frac{\text{x}^2+2\text{x}+1}{\text{x}(\text{x}+2)}}\Bigg)=\frac{\pi}{12}$
$\Rightarrow\tan^{-1}\Big(\frac{2}{\text{x}^2+2\text{x}+1}\Big)=\frac{\pi}{12}$
$\Rightarrow\Big(\frac{2}{\text{x}^2+2\text{x}+1}\Big)=\tan\frac{\pi}{12}$
$\Rightarrow\Big(\frac{2}{\text{x}^2+2\text{x}+1}\Big)=\tan\Big(\frac{\pi}{3}-\frac{\pi}{4}\Big)$
$\Rightarrow\Big(\frac{2}{\text{x}^2+2\text{x}+1}\Big)=\frac{\tan\frac{\pi}{3}+\tan\frac{\pi}{4}}{1+\tan\frac{\pi}{3}\times\tan\frac{\pi}{4}}$
$\Rightarrow\Big(\frac{2}{\text{x}^2+2\text{x}+1}\Big)=\frac{\sqrt3-1}{\sqrt3+1}$
$\Rightarrow\Big(\frac{2}{\text{x}^2+2\text{x}+1}\Big)=\frac{\sqrt3-1}{\sqrt3+1}\times\frac{\sqrt3+1}{\sqrt3+1}$
$\Rightarrow\Big(\frac{2}{\text{x}^2+2\text{x}+1}\Big)=\frac{2}{\big(\sqrt3+1\big)^2}$
$\Rightarrow\frac{1}{(\text{x}+1)^2}\frac{1}{\big(\sqrt3+1\big)^2}$
$\Rightarrow\text{x}+1=\sqrt3+1$
$\Rightarrow\text{x}=\sqrt3$
View full question & answer→Question 105 Marks
Prove the following results:
$\sin^{-1}\frac{12}{13}+\cos^{-1}\frac{4}{5}+\tan^{-1}\frac{63}{16}=\pi$
Answer$\sin^{-1}\Big(\frac{12}{13}\Big)+\cos^{-1}\Big(\frac{4}{5}\Big)+\tan^{-1}\Big(\frac{63}{16}\Big)=\pi$ $\text{L.H.S}=\sin^{-1}\Big(\frac{12}{13}\Big)+\cos^{-1}\Big(\frac{4}{5}\Big)+\tan^{-1}\Big(\frac{63}{16}\Big)$ $=\tan^{-1}\begin{bmatrix}\frac{\frac{12}{13}}{\sqrt{1-\big(\frac{12}{13}\big)^2}}\end{bmatrix}+\tan^{-1}\begin{bmatrix}\frac{\sqrt{1-\big(\frac{4}{5}\big)^2}}{\frac{4}{5}}\end{bmatrix}+\tan^{-1}\Big(\frac{63}{16}\Big)$ $\bigg\{\text{Since}\sin^{-1}\text{x}=\tan^{-1}\bigg(\frac{\text{x}}{\sqrt{1-\text{x}^2}}\bigg)\cos^{-1}\text{x}=\tan^{-1}\bigg(\frac{\sqrt{1-\text{x}^2}}{\text{x}}\bigg)\bigg\}$ $=\tan^{-1}\Bigg(\frac{\frac{12}{13}}{\frac{5}{13}}\Bigg)+\tan^{-1}\Bigg(\frac{\frac{3}{5}}{\frac{4}{5}}\Bigg)+\tan^{-1}\Big(\frac{63}{16}\Big)$ $=\tan^{-1}\Big(\frac{12}{5}\Big)+\tan^{-1}\Big(\frac{3}{4}\Big)+\tan^{-1}\Big(\frac{63}{16}\Big)$ $=\pi+\tan^{-1}\Bigg(\frac{\frac{12}{5}+\frac{3}{4}}{1-\frac{12}{5}\times\frac{3}{4}}\Bigg)+\tan^{-1}\Big(\frac{63}{16}\Big)$ $\Big\{\text{Since}\tan^{-1}\text{x}+\tan^{-1}\text{y}=\pi+\tan^{-1}\Big(\frac{\text{x+y}}{1-\text{xy}}\Big)\\\text{ if }\text{x}>0,\text{y}>0\text{ and }\text{xy}>0\Big\}$ $=\pi+\tan^{-1}\Bigg(\frac{\frac{63}{20}}{-\frac{16}{20}}\Bigg)+\tan^{-1}\Big(\frac{63}{16}\Big)$ $=\pi+\tan^{-1}\Big(-\frac{63}{16}\Big)+\tan^{-1}\Big(\frac{63}{16}\Big)$ $=\pi-\tan^{-1}\Big(\frac{63}{16}\Big)+\tan^{-1}\Big(\frac{63}{16}\Big)$ $\Big\{\text{Since}\tan^{-1}(-\text{x})=\tan^{-1}\text{x}\Big\}$ $=\pi$ Hence, $\sin^{-1}\Big(\frac{12}{13}\Big)+\cos^{-1}\Big(\frac{4}{5}\Big)+\tan^{-1}\Big(\frac{63}{16}\Big)=\pi$
View full question & answer→Question 115 Marks
Solve the following equation for x:
$\tan^{-1}\frac{\text{x}-2}{\text{x}-1}+\tan^{-1}\frac{\text{x}+2}{\text{x}+1}=\frac{\pi}{4}$
Answer$\tan^{-1}\frac{\text{x}-2}{\text{x}-1}+\tan^{-1}\frac{\text{x}+2}{\text{x}+1}=\frac{\pi}{4}$
$\Rightarrow\tan^{-1}\Bigg(\frac{\big(\frac{\text{x}-2}{\text{x}-1}\big)+\big(\frac{\text{x}+2}{\text{x}+1}\big)}{1-\big(\frac{\text{x}-2}{\text{x}-1}\big)\big(\frac{\text{x}+2}{\text{x}+1}\big)}\Bigg)=\frac{\pi}{4}$
$\Big[ \tan^{-1}\text{x}+\tan^{-1}\text{y}=\tan^{-1}\Big(\frac{\text{x}+\text{y}}{1-\text{xy}}\Big)\Big]$
$\Rightarrow\frac{\Big(\frac{(\text{x}-2)(\text{x}+1)+(\text{x}-1(\text{x}+2))}{(\text{x}-1)(\text{x}+1)}\Big)}{\Big(\frac{(\text{x}-1)(\text{x}+1)-(\text{x}-2)(\text{x}+2)}{(\text{x}-1)(\text{x}+1)}\Big)}=\tan\frac{\pi}{4}$
$\Rightarrow\frac{(\text{x}-2)(\text{x}+1)+(\text{x}-1)(\text{x}+2)}{(\text{x}-1)(\text{x}+1)+(\text{x}-2)(\text{x}+2)}=1$
$\Rightarrow\frac{2\text{x}^2-4}{3}=1$
$\Rightarrow2\text{x}^2-4=3$
$\Rightarrow2\text{x}^2=7$
$\Rightarrow\text{x}^2=\frac{7}{2}$
$\therefore\ \Rightarrow\text{x}=\pm\sqrt{\frac{7}{2}}$
View full question & answer→Question 125 Marks
Solve the following equation for x:
$\tan^{-1}\Big(\frac{\text{x}-2}{\text{x}-1}\Big)+\tan^{-1}\Big(\frac{\text{x}+2}{\text{x}+1}\Big)=\frac{\pi}{4}$
Answer$\tan^{-1}\Big(\frac{\text{x}-2}{\text{x}-1}\Big)+\tan^{-1}\Big(\frac{\text{x}+2}{\text{x}+1}\Big)=\frac{\pi}{4}$
$\Rightarrow\tan^{-1}\Big(\frac{\text{x}-2}{\text{x}-1}\Big)+\tan^{-1}\Big(\frac{\text{x}+2}{\text{x}+1}\Big)=\tan^{-1}1$
$\Rightarrow\tan^{-1}\Big(\frac{\text{x}-2}{\text{x}-1}\Big)=\tan^{-1}1-\tan^{-1}\Big(\frac{\text{x}+2}{\text{x}+1}\Big)$
$\Rightarrow\tan^{-1}\Big(\frac{\text{x}-2}{\text{x}-1}\Big)=\tan^{-1}\Bigg(\frac{1-\frac{\text{x}+2}{\text{x}+1}}{1+\frac{\text{x}+2}{\text{x}+1}}\Bigg)$
$\Rightarrow\tan^{-1}\Big(\frac{\text{x}-2}{\text{x}-1}\Big)=\tan^{-1}\Big(\frac{\text{x}+1-\text{x}-2}{\text{x}+1+\text{x}+2}\Big)$
$\Rightarrow\tan^{-1}\Big(\frac{\text{x}-2}{\text{x}-1}\Big)=\tan^{-1}\Big(\frac{-1}{2\text{x}+3}\Big)$
$\Rightarrow\frac{\text{x}-2}{\text{x}-1}=\frac{-1}{2\text{x}+3}$
$\Rightarrow2\text{x}^2+3\text{x}-4\text{x}-6=-\text{x}+1$
$\Rightarrow2\text{x}^2=1+6$
$\Rightarrow\text{x}^2=7$
$\Rightarrow\text{x}=\pm\sqrt{\frac{7}{2}}$
View full question & answer→Question 135 Marks
Evaluate the following:
$\sin\Big(\sec^{-1}\frac{17}{8}\Big)$
Answer$\sin\Big(\sec^{-1}\frac{17}{8}\Big)$
$=\sin\Big(\cos^{-1}\frac{8}{17}\Big)$
$=\sin\Bigg[\sin^{-1}\sqrt{1-\Big(\frac{8}{17}\Big)^2}\Bigg]$ $\Big[{\therefore\ \cos^{-1}}\text{x}=\sin^{-1}\sqrt{1-\text{x}^2}\Big]$
$=\sin\Bigg[\sin^{-1}\Bigg(\sqrt{1-\frac{64}{289}}\Bigg)\Bigg]$
$=\sin\Bigg[\sin^{-1}\Bigg(\sqrt{\frac{225}{289}}\Bigg)\Bigg]$
$=\sin\Big[\sin^{-1}\frac{15}{17}\Big]$
$=\frac{15}{17}$
View full question & answer→Question 145 Marks
Solve: $\cos\Big\{2\sin^{-1}\{-\text{x}\}\Big\}=0$
Answer$0=\cos\Big\{2\sin^{-1}\{-\text{x}\}\Big\}$
$0=\cos\Big(\sin^{-1}\Big(-2\times\sqrt{1-\text{x}^2}\Big)\Big)\\.....\Big[2\sin^{-1}\text{x}=\sin^{-1}\Big(2\times\sqrt{1-\text{x}^2}\Big)\Big ]$
$$$0=\cos\Big(\cos^{-1}\sqrt{1-\big(4\text{x}^2-4\text{x}^4}\big)\Big)\\.....\Big[\sin^{-1}\text{x}=\cos^{-1}\sqrt{1-\text{x}^2}\Big]$
$$$0=\sqrt{1-\big(4\text{x}^2-4\text{x}^4\big)}$
$0=1-\big(4\text{x}^2-4\text{x}^4\big)$
$4\text{x}^4-4\text{x}^2+1=0$
$4\text{x}^4-2\text{x}^2-2\text{x}^2+1=0$
$2\text{x}^2\big(2\text{x}^2-1\big)-1\big(2\text{x}^2-1\big)=0$
$\big(2\text{x}^2-1\big)^2=0$
$2\text{x}^2-1=0$
$\text{x}=\pm\frac{1}{\sqrt2}$
View full question & answer→Question 155 Marks
Write the following in the simplest form:
$\tan^{-1}\sqrt{\frac{\text{x}}{\text{a}+\sqrt{\text{a}^2-\text{x}^2}}},-\text{a}<\text{x}<\text{a}$
Answer$\tan^{-1}\sqrt{\frac{\text{x}}{\text{a}+\sqrt{\text{a}^2-\text{x}^2}}},-\text{a}<\text{x}<\text{a}$
Let, $\text{x}=\text{a}\sin\theta$
$\tan^{-1}\bigg\{\frac{\text{a}\sin\theta}{1+\sqrt{\text{a}^2-\text{a}^2\sin^2\theta}}\bigg\}$
$=\tan^{-1}\Bigg\{\sqrt{\frac{\text{a}\sin\theta}{\text{a}+\text{a}\sqrt{1-\sin^2\theta}}}\Bigg\}$
$=\tan^{-1}\Big\{\frac{\text{a}\sin\theta}{\text{a}(1+\cos\theta)}\Big\}$ $\big\{\text{Since},1-\sin^2\theta=\cos^2\theta\big\}$
$=\tan^{-1}\Big\{\frac{\sin\theta+1}{1+\cos\theta}\Big\}$
$=\tan^{-1}\Bigg\{\frac{\frac{2\sin\theta}{2}\frac{\cos\theta}{2}}{\frac{2\cos^2\theta}{2}}\Bigg\}$ $\Big\{\text{Since},\sin\theta=\frac{2\sin\theta}{2}\frac{\cos\theta}{2},1+\cos\theta=\frac{2\cos^2\theta}{2}\Big\}$
$=\tan^{-1}\bigg\{\frac{\frac{\sin\theta}{2}}{\frac{\cos\theta}{2}}\bigg\}$
$=\tan^{-1}\Big\{\frac{\tan\theta}{2}\Big\}$ $\Big\{\text{Since},\frac{\sin\theta}{\cos\theta}=\tan\theta\Big\}$
$=\frac{\theta}{2}$
$=\frac{1}{2}\sin^{-1}\text{x}$ $\Big\{\text{Since},\text{x}=\sin\theta\Rightarrow\theta=\sin^{-1}\Big(\frac{\text{x}}{\text{a}}\Big)\Big\}$
View full question & answer→Question 165 Marks
Evaluate the following:
$\tan\frac{1}{2}\Big(\cos^{-1}\frac{\sqrt5}{3}\Big)$
AnswerLet $\frac{1}{2}\sin^{-1}\frac{3}{4}=\text{x}$
$ \sin^{-1}\frac{3}{4}=2\text{x}$
$ \sin2\text{x}=\frac{3}{4}$
$\cos2\text{x}=\frac{\sqrt7}{4}$
$\tan\Big(\frac{1}{2}\sin^{-1}\frac{3}{4}\Big)$
$=\tan\text{x}$
$ =\sqrt{\frac{1-\cos2\text{x}}{1+\cos2\text{x}}}$
$=\sqrt{\frac{1-\frac{\sqrt7}{4}}{1+\frac{\sqrt7}{4}}}$
$=\sqrt{\frac{4-\sqrt7}{4+\sqrt7}}$
$=\sqrt{\frac{\big(4-\sqrt7\big)\big(4-\sqrt7\big)}{\big(4+\sqrt7\big)\big(4-\sqrt7\big)}}$
$=\sqrt{\frac{\big(4-\sqrt7\big)^2}{9}}$
$ =\frac{4-\sqrt7}{3}$
View full question & answer→Question 175 Marks
Solve the following equation for x:
$\tan^{-1}2\text{x}+\tan^{-1}3\text{x}=\text{n}\pi+\frac{3\pi}{4}$
AnswerGiven$\tan^{-1}2\text{x}+\tan^{-1}3\text{x}=\text{n}\pi+\frac{3\pi}{4}....(1)$
$\Rightarrow\tan^{-1}\Big(\frac{2\text{x}+3\text{x}}{1-2\text{x}\times3\text{x}}\Big)=\text{n}\pi+\frac{3\pi}{4}$
$\Big\{\text{Since}\tan^{-1}\text{x}+\tan^{-1}\text{y}=\tan^{-1}\Big(\frac{\text{x+y}}{1-\text{xy}}\Big)\text{ if }\text{xy}<1\Big\}$
$\Rightarrow\tan^{-1}\Big(\frac{5\text{x}}{1-6\text{x}^2}\Big)=\text{n}\pi+\frac{3\pi}{4},6\text{x}^2<1$
$\Rightarrow\frac{5\text{x}}{1-6\text{x}^2}=\tan\Big(\text{n}\pi+\frac{3\pi}{4}\Big),6\text{x}^2<1$
$\Rightarrow\frac{5\text{x}}{1-6\text{x}^2}-1,6\text{x}^2<1$
$\Rightarrow5\text{x}=-1+6\text{x}^2,6\text{x}^2<1$
$\Rightarrow6\text{x}^2-5\text{x}-1=0,\text{x}^2<\frac{1}{6}$
$\Rightarrow6\text{x}^2-6\text{x}+\text{x}-1=0,-\frac{1}{\sqrt6}<\text{x}<\frac{1}{6}$
$\Rightarrow6\text{x}(\text{x}-1)+1(\text{x}-1)=0,-\frac{1}{\sqrt6}<\text{x}<\frac{1}{6}$
$\Rightarrow(6\text{x}+1)(\text{x}-1)=0$
$\Rightarrow6\text{x}+1=0$ or $\text{x}-1=0$
$\Rightarrow\text{x}=-\frac{1}{6}$ or $\text{x}=1$
Since, $\text{x}=1\notin\Big(-\frac{1}{\sqrt6},\frac{1}{\sqrt6}\Big)$
So,
x = 1 is not root of the given equation (i).
Since,
$\text{x}=1\in\Big(-\frac{1}{\sqrt6},\frac{1}{\sqrt6}\Big)$
So,
$\text{x}=-\frac{1}{6}$ is the root of the given equation (i).
Hence,
$\text{x}=-\frac{1}{6}$
View full question & answer→Question 185 Marks
Evaluate the following:
$\tan\Big\{2\tan^{-1}\frac{1}{5}-\frac{\pi}{4}\Big\}$
Answer$\tan\Big(2\tan^{-1}\frac{1}{5}-\frac{\pi}{4}\Big)=\tan\Big(2\tan^{-1}\frac{1}{5}-\tan^{-1}1\Big)$
$=\tan\begin{bmatrix}\tan^{-1}\begin{Bmatrix}\frac{2\times\frac{1}{5}}{1-\Big(\frac{1}{5}\Big)^2}\end{Bmatrix}-\tan^{-1}1\end{bmatrix}$
$ \Big[\because\ 2\tan^{-1}\text{x}=\tan^{-1}\Big\{\frac{2\text{x}}{1-\text{x}^2}\Big\}\Big]$
$=\tan\Bigg[\tan^{-1}\Bigg\{\frac{\frac{2}{5}}{\frac{24}{25}}\Bigg\}-\tan^{-1}1\Bigg]$
$ =\tan\Big[\tan^{-1}\frac{5}{12}+\tan^{-1}1\Big]$
$=\tan^{-1}\Bigg[\tan^{-1}\Bigg(\frac{\frac{15}{12}-1}{1+\frac{5}{12}}\Bigg)\Bigg]$
$\Big[\because\tan^{-1}\text{x}-\tan^{-1}\text{y}=\tan^{-1}\Big(\frac{\text{x}-\text{y}}{1+\text{xy}}\Big)\Big]$
$ =\tan\Bigg[\tan^{-1}\Bigg(\frac{\frac{-7}{12}}{\frac{17}{12}}\Bigg)\Bigg]$
$=\tan\Big[\tan^{-1}\frac{-7}{17}\Big]$
$=\frac{-7}{17}$
View full question & answer→Question 195 Marks
Prove the following results:
$2\sin^{-1}\frac{3}{5}=\tan^{-1}\frac{24}{7}$
Answer$2\sin^{-1}\frac{3}{5}=\tan^{-1}\frac{24}{7}$ $\text{L.H.S}=2\sin^{-1}\frac{3}{5}$ $=2\times\tan^{-1}\begin{bmatrix}\frac{\frac{3}{5}}{\sqrt{1-\Big(\frac{3}{5}\Big)^2}}\end{bmatrix}$ $\bigg\{\text{Since }\sin^{-1}\text{x}=\tan^{-1}\frac{\text{x}}{\sqrt{1-\text{x}^2}}\bigg\}$ $=2\tan^{-1}\Bigg(\frac{\frac{3}{5}}{\frac{4}{5}}\Bigg)$ $=2\tan^{-1}\Big(\frac{3}{4}\Big)$ $=\tan^{-1}\Bigg(\frac{2\times\frac{3}{4}}{1-\big(\frac{3}{4}\big)^2}\Bigg)$ $\Big\{\text{Since }2\tan^{-1}\text{x}=\tan^{-1}\frac{2\text{x}}{1-\text{x}^2}\Big\}$ $=\tan^{-1}\bigg(\frac{\frac{3}{2}}{\frac{7}{16}}\bigg)$ $=\tan^{-1}\Big(\frac{24}{7}\Big)$ $=\text{R.H.S}$So,
$2\sin^{-1}\frac{3}{5}=\tan^{-1}\frac{24}{7}$
View full question & answer→Question 205 Marks
Prove that
$\text{L.H.S}=\sin\Big\{\tan^{-1}\frac{1-\text{x}^2}{2\text{x}}+\cos^{-1}\frac{1-\text{x}^2}{1+\text{x}^2}\Big\}=1$
Answer$\text{L.H.S}=\sin\Big[\tan^{-1}\Big(\frac{1-\text{x}^2}{2\text{x}}\Big)+\cos^{-1}\Big(\frac{1-\text{x}^2}{1+\text{x}^2}\Big)\Big]=1$
$\text{L.H.S}=\sin\Big[\tan^{-1}\Big(\frac{1-\text{x}^2}{2\text{x}}\Big)+\cos^{-1}\Big(\frac{1-\text{x}^2}{1+\text{x}^2}\Big)\Big]$
$=\sin\Big[\tan^{-1}\Big(\frac{1-\text{x}^2}{2\text{x}}\Big)+2\tan^{-1}\text{x}\Big]$
$\Big\{\text{Since }2\tan^{-1}\text{x}=\cos^{-1}\Big(\frac{1-\text{x}^2}{1+\text{x}^2}\Big)\Big\}$
$=\sin\big[\tan^{-1}\big]\Big(\frac{1-\text{x}^2}{2\text{x}}\Big)+\tan^{-1}\Big(\frac{2\text{x}}{1-\text{x}^2}\Big)$
$\Big\{\text{Since }2\tan^{-1}\text{x}=\tan\frac{2\text{x}}{1-\text{x}^2}\Big\}$
$=\sin\begin{bmatrix}\tan^{-1}\begin{pmatrix}\frac{\frac{1-\text{x}^2}{2\text{x}}+\frac{2\text{x}}{1-\text{x}^2}}{1-\frac{1-\text{x}^2}{2\text{x}}\times\frac{2\text{x}}{1-\text{x}^2}}\end{pmatrix}\end{bmatrix}$
$\Big\{\text{Since }\tan^{-1}\text{x}+\tan^{-1}\text{y}=\tan^{-1}\Big(\frac{\text{x}+\text{y}}{1-\text{xy}}\Big)\Big\}$
$=\sin\Bigg[\tan^{-1}\Bigg(\frac{\frac{1+\text{x}^4-2\text{x}^2+4\text{x}^2}{2\text{x}(1-\text{x}^2)}}{0}\Bigg)\Bigg]$
$=\sin\big[\tan^{-1}(\infty)\big]$
$=\sin\Big[\frac{\pi}{2}\Big]$
$=1$
$=\text{R.H.S}$
Hence,
$\text{L.H.S}=\sin\Big[\tan^{-1}\Big(\frac{1-\text{x}^2}{2\text{x}}\Big)+\cos^{-1}\Big(\frac{1-\text{x}^2}{1+\text{x}^2}\Big)\Big]=1$
View full question & answer→Question 215 Marks
Solve the following equation for x:
$\tan^{-1}\frac{\pi}{2}+\tan^{-1}\frac{\pi}{3}=\frac{\pi}{4},0<\text{x}<\sqrt6$
AnswerGiven,
$\tan^{-1}\frac{\pi}{2}+\tan^{-1}\frac{\pi}{3}=\frac{\pi}{4},0<\text{x}<\sqrt6$
$\Rightarrow\tan^{-1}\bigg[\frac{\frac{\text{x}}{2}+\frac{\text{x}}{3}}{1-\frac{\text{x}}{2}\times\frac{\text{x}}{3}}\bigg]=\frac{\pi}{4}$
$\Big\{\text{Since},\tan^{-1}\text{x}+\tan^{-1}\text{y}=\tan^{-1}\Big(\frac{\text{x}+\text{y}}{1-\text{xy}}\Big)\Big\}$
$\Rightarrow\tan^{-1}\Bigg[\frac{\frac{5\text{x}}{6}}{\frac{(6-\text{x}^2)}{16}}\Bigg]=\frac{\pi}{4}$
$\Rightarrow\tan^{-1}\Big[\frac{5\text{x}}{6-\text{x}^2}\Big]=\frac{\pi}{4}$
$\Rightarrow\frac{5\text{x}}{6-\text{x}^2}=\tan\frac{\pi}{4}$
$\Rightarrow\frac{5\text{x}}{6-\text{x}^2}=1$
$\Rightarrow5\text{x}=6\text{x}^2$
$\Rightarrow\text{x}^2+5\text{x}-6=0$
$\Rightarrow\text{x}^2+6\text{x}-\text{x}-6=0$
$\Rightarrow\text{x}(\text{x}+6)-1(\text{x}+6)=0$
$\Rightarrow(\text{x}+6)(\text{x}-1)=0$
$\Rightarrow\text{x}=-6\text{ or }\text{x}=1$
But, $0<\text{x}<\sqrt6,$ so,
$\text{x}=1$
View full question & answer→Question 225 Marks
Solve the following:
$\sin^{-1}\text{x}+\sin^{-1}2\text{x}=\frac{\pi}{3}$
AnswerWe know $\sin^{-1}\text{x}+\sin^{-1}\text{y}=\sin^{-1}\Big[\text{x}\sqrt{1-\text{y}^2}+\text{y}\sqrt{1-\text{x}^2}\Big]$ $\therefore\ \sin^{-1}\text{x}+\sin^{-1}2\text{x}=\frac{\pi}{3}$ $\Rightarrow\sin^{-1}\text{x}+\sin^{-1}2\text{x}=\sin^{-1}\Big(\frac{\sqrt{3}}{2}\Big)$$\Rightarrow\sin^{-1}\text{x}-\sin^{-1}\Big(\frac{\sqrt{3}}{2}\Big)=-\sin^{-1}2\text{x}$
$\Rightarrow\sin^{-1}\bigg[\text{x}\sqrt{1-\frac{3}{4}}+\frac{\sqrt3}{2}\sqrt{1-\text{x}^2}\bigg]=-\sin^{-1}2\text{x}$
$\Rightarrow\sin^{-1}\Big[\frac{\text{x}}{2}+\frac{\sqrt3}{2}\sqrt{1-\text{x}^2}\Big]=\sin^{-1}(-2\text{x})$
$\Rightarrow\frac{\text{x}}{2}+\frac{\sqrt3}{2}\sqrt{1-\text{x}^2}=-2\text{x}$
$\Rightarrow5\text{x}=-\sqrt3\sqrt{1-\text{x}^2}$
Squaring both the sides, $25\text{x}^2=3-3\text{x}^2$ $\Rightarrow28\text{x}^2=3$ $\Rightarrow\text{x}=\pm\frac{1}{2}\sqrt{\frac{3}{7}}$
View full question & answer→Question 235 Marks
Evaluate: $\sin\Big\{\cos^{-1}\Big(-\frac{3}{5}\Big)+\cot^{-1}\Big(-\frac{5}{12}\Big)\Big\}$
Answer$\sin\Big\{\cos^{-1}\Big(-\frac{3}{5}\Big)+\cot^{-1}\Big(-\frac{5}{12}\Big)\Big\}$
$=\sin\Big\{\pi-\cos^{-1}\Big(\frac{3}{5}\Big)+\pi-\cot^{-1}\Big(\frac{5}{12}\Big)\Big\}$
$=\sin\Big\{2\pi-\Big[\cos^{-1}\Big(\frac{3}{5}\Big)+\cot^{-1}\Big(\frac{5}{12}\Big)\Big]\Big\}$
$=-\sin\Big\{\cos^{-1}\Big(\frac{3}{5}\Big)+\cot^{-1}\Big(\frac{5}{12}\Big)\Big\}$
$=-\sin\begin{Bmatrix}\sin^{-1}\Bigg[\sqrt{1-\Big(\frac{3}{5}\Big)^2}\Bigg]=\sin^{-1}\begin{bmatrix}\frac{\frac{12}{5}}{\sqrt{1+\big(\frac{12}{5}\big)^2}}\end{bmatrix}\end{Bmatrix}$
$=-\sin\Big(\sin^{-1}\frac{4}{5}+\sin^{-1}\frac{12}{13}\Big)$
$=-\sin\Bigg\{\sin^{-1}\Bigg[\frac{4}{5}\times\sqrt{1-\Big(\frac{12}{13}\Big)^2}+\frac{12}{13}\times\sqrt{1-\Big(\frac{4}{5}\Big)^2}\Bigg]\Bigg\}$
$=-\sin\Big[\sin^{-1}\Big(\frac{20}{65}+\frac{36}{64}\Big)\Big]$
$=-\sin\Big[\sin^{-1}\Big(\frac{56}{65}\Big)\Big]$
$=-\frac{56}{65}$
View full question & answer→Question 245 Marks
Prove the following results:
$2\tan^{-1}\Big(\frac{1}{2}\Big)+\tan^{-1}\Big(\frac{1}{7}\Big)=\tan^{-1}\Big(\frac{31}{17}\Big)$
Answer$\text{L.H.S}=2\tan^{-1}\frac{1}{2}+\tan^{-1}\frac{1}{7}$ $=\tan^{-1}\Bigg\{\frac{2\times\frac{1}{2}}{1-\big(\frac{1}{2}\big)^2}\Bigg\}+\tan^{-1}\frac{1}{7}$ $\Big[\because\ 2\tan^{-1}\text{x}=\tan^{-1}\Big\{\frac{2\text{x}}{1-\text{x}^2}\Big\}\Big]$ $=\tan^{-1}\Bigg\{\frac{1}{\frac{3}{4}}\Bigg\}+\tan^{-1}\frac{1}{7}$ $=\tan^{-1}\frac{4}{3}+\tan^{-1}\frac{1}{7}$ $=\tan^{-1}\Bigg(\frac{\frac{4}{3}+\frac{1}{7}}{1-\frac{4}{3}\times\frac{1}{7}}\Bigg)$ $\Big[\because\ \tan^{-1}\text{x}+\tan^{-1}\text{y}=\tan^{-1}\Big(\frac{\text{x}+\text{y}}{1-\text{xy}}\Big)\Big]$$=\tan^{-1}\Bigg(\frac{\frac{31}{21}}{\frac{17}{21}}\Bigg)$
$=\tan^{-1}\frac{31}{17}=\text{R.H.S}$
View full question & answer→Question 255 Marks
Solve $\cos^{-1}\sqrt3\text{x}+\cos^{-1}\text{x}=\frac{\pi}{2}$
Answer$\cos^{-1}\sqrt3\text{x}+\cos^{-1}\text{x}=\frac{\pi}{2}$
$ \Rightarrow\cos^{-1}\Big[\sqrt3\text{x}\times\text{x}-\sqrt{1-\big(\sqrt3\text{x}\big)^2}\sqrt{1-\text{x}^2}\Big]=\frac{\pi}{2}$
$\Big[\because\ \cos^{-1}\text{x}+\cos^{-1}\text{y}=\cos^{-1}\Big(\text{xy}-\sqrt{1-\text{x}^2}\sqrt{1-\text{y}^2}\Big)\Big]$
$ \Rightarrow\cos^{-1}\Big[\sqrt3\text{x}^2-\sqrt{1-3\text{x}^2}\sqrt{1-\text{x}^2}\Big]=\frac{\pi}{2}$
$\Rightarrow\sqrt3\text{x}^2-\sqrt{1-3\text{x}^2}\sqrt{1-\text{x}^2}=\cos\frac{\pi}{2}$
$ \Rightarrow\sqrt3\text{x}^2=\sqrt{1-3\text{x}^2}\sqrt{1-\text{x}^2}$
$ \Rightarrow3\text{x}^2=\big(1-3\text{x}^2\big)\big(1-\text{x}^2\big)$
$ \Rightarrow3\text{x}^4=1-3\text{x}^2+3\text{x}^4-\text{x}^2$
$ \Rightarrow4\text{x}^2=1$
$ \Rightarrow\text{x}^2=\frac{1}{4}$
$ \Rightarrow\text{x}=\pm\frac{1}{2}$
View full question & answer→Question 265 Marks
If $\sin^{-1}\text{x}+\sin^{-1}\text{y}+\sin^{-1}\text{z}+\sin^{-1}\text{t}=2\pi,$ then find the value of $x^2 + y^2 + z^2 + t^2$
AnswerWe know that the maximum value of $\sin^{-1}\text{x}+\sin^{-1}\text{y}+\sin^{-1}\text{z}$ and $\sin^{-1}\text{t}$ is $\frac{\pi}{2}$ Now,$\text{L.H.S}=\sin^{-1}\text{x}+\sin^{-1}\text{y}+\sin^{-1}\text{z}+\sin^{-1}\text{t}$
$=\frac{\pi}{2}+\frac{\pi}{2}+\frac{\pi}{2}+\frac{\pi}{2}$
$=2\pi=\text{R.H.S}$
Now, $\sin^{-1}\text{x}=\frac{\pi}{2},\sin^{-1}\text{y}=\frac{\pi}{2},\sin^{-1}\text{z}=\frac{\pi}{2}$ and $\sin^{-1}\text{t}=\frac{\pi}{2}$
$\Rightarrow\text{x}=\sin\frac{\pi}{2},\text{y}=\sin\frac{\pi}{2},\text{z}=\sin\frac{\pi}{2}$ and $\text{t}=\sin\frac{\pi}{2}$
$\Rightarrow x = 1, y = 1, z= 1$ and $t = 1$
$\therefore$ $x^2 + y^2 + z^2 + t^2 = 1 + 1 + 1 + 1 = 4$
View full question & answer→Question 275 Marks
Solve the equation
$\cos^{-1}\frac{\text{a}}{\text{x}}-\cos^{-1}\frac{\text{b}}{\text{x}}=\cos^{-1}\frac{1}{\text{b}}-\cos^{-1}\frac{1}{\text{b}}-\cos^{-1}\frac{1}{\text{a}}$
Answer$\cos^{-1}\frac{\text{a}}{\text{x}}-\cos^{-1}\frac{\text{b}}{\text{x}}=\cos^{-1}\frac{1}{\text{b}}-\cos^{-1}\frac{1}{\text{b}}-\cos^{-1}\frac{1}{\text{a}}$
$\cos^{-1}\frac{\text{a}}{\text{x}}-\cos^{-1}\frac{\text{1}}{\text{a}}=\cos^{-1}\frac{\text{b}}{\text{x}}-\cos^{-1}\frac{\text{1}}{\text{b}}$
$\cos^{-1}\bigg[\frac{1}{\text{x}}-\sqrt{1-\frac{\text{a}^2}{\text{x}^2}}\sqrt{1-\frac{1}{\text{a}^2}}\bigg]$ $=\cos^{-1}\bigg[\frac{1}{\text{x}}-\sqrt{1-\frac{\text{b}^2}{\text{x}^2}}\sqrt{1-\frac{1}{\text{b}^2}}\bigg] $
$\Rightarrow\frac{1}{\text{x}}-\sqrt{1-\frac{\text{a}^2}{\text{x}^2}}\sqrt{1-\frac{1}{\text{a}^2}}=\frac{1}{\text{x}}-\sqrt{1-\frac{\text{b}^2}{\text{x}^2}}\sqrt{1-\frac{1}{\text{b}^2}} $
$\Rightarrow\sqrt{1-\frac{\text{a}^2}{\text{x}^2}}\sqrt{1-\frac{\text{1}}{\text{a}^2}}=\sqrt{1-\frac{\text{b}^2}{\text{x}^2}}\sqrt{1-\frac{\text{1}}{\text{b}^2}}$
$\Rightarrow \Big(1-\frac{\text{a}^2}{\text{x}^2}\Big)\Big(1-\frac{1}{\text{a}^2}\Big)=\Big(1-\frac{\text{b}^2}{\text{x}^2}\Big)\Big(1-\frac{\text{1}}{\text{b}^2}\Big)$
$\Rightarrow1-\frac{1}{\text{a}^2}-\frac{\text{a}^2}{\text{x}^2}+\frac{1}{\text{x}^2}=1-\frac{1}{\text{b}^2}-\frac{\text{a}^2}{\text{x}^2}+\frac{1}{\text{x}^2}$
$\Rightarrow\frac{\text{b}^2}{\text{x}^2}-\frac{\text{a}^2}{\text{x}^2}=\frac{1}{\text{a}^2}-\frac{1}{\text{b}^2}$
$\Rightarrow\big(\text{b}^2-\text{a}^2\big)\text{a}^2\text{b}^2=\text{x}^2\big(\text{b}^2-\text{a}^2\big) $
$\Rightarrow\text{x}^2=\text{a}^2\text{b}^2$
$\Rightarrow\text{x}=\text{ab}$
View full question & answer→Question 285 Marks
Prove the following results:
$\tan^{-1}\frac{1}{4}+\tan^{-1}\frac{2}{9}=\frac{1}{2}\cos^{-1}\frac{3}{5}=\frac{1}{2}\sin^{-1}\Big(\frac{4}{5}\Big)$
Answer$\text{L.H.S}=\tan^{-1}\frac{1}{4}+\tan^{-1}\frac{2}{9}$
$=\tan^{1}\Bigg(\frac{\frac{1}{4}+\frac{2}{9}}{1-\frac{1}{4}\times\frac{2}{9}}\Bigg)$ $ \Big[\because\ \tan^{-1}\text{x}+\tan^{-1}\text{y}=\tan^{-1}\Big(\frac{\text{x}+\text{y}}{1-\text{xy}}\Big)\Big]$
$ =\tan^{-1}\bigg(\frac{\frac{17}{36}}{\frac{34}{36}}\bigg)$
$ =\tan^{-1}\frac{1}{2}$
$=\frac{1}{2}\cos^{-1}\Bigg(\frac{1-\frac{1}{4}}{1+\frac{1}{4}}\Bigg)$ $\Big[\because\ \tan^{-1}\text{x}=\frac{1}{2}\cos^{-1}\Big(\frac{1-\text{x}^2}{1+\text{x}^2}\Big)\Big]$
$=\frac{1}{2}\cos^{-1}\Bigg(\frac{\frac{3}{4}}{\frac{5}{4}}\Bigg)$
$=\frac{1}{2}\cos^{-1}\Big(\frac{3}{5}\Big)$
Now,
$\tan^{-1}\frac{1}{2}=\frac{1}{2}\sin^{-1}\Bigg(\frac{\frac{2}{2}}{1+\frac{1}{4}}\Bigg)$ $\Big[\because\ \tan^{-1}\text{x}=\frac{1}{2}\sin^{-1}\Big(\frac{2\text{x}}{1+\text{x}^2}\Big)\Big]$
$=\frac{1}{2}\sin^{-1}\bigg(\frac{1}{\frac{5}{4}}\bigg)$
$=\frac{1}{2}\sin^{-1}\Big(\frac{4}{5}\Big)$
View full question & answer→Question 295 Marks
If $\sin^{-1}\frac{2\text{a}}{1+\text{a}^2}-\cos^{-1}\frac{1-\text{b}^2}{1+\text{b}^2}=\tan^{-1}\frac{2\text{x}}{1-\text{x}^2},$ then prove that $\text{x}=\frac{\text{a}-\text{b}}{1+\text{ab}}$
AnswerLet:
$\text{a}=\tan\text{m}$
$\text{b}=\tan\text{n}$
$\text{x}=\tan\text{y}$
Now,
$\sin^{-1}\frac{2\text{a}}{1+\text{a}^2}-\cos^{-1}\frac{1-\text{b}^2}{1+\text{b}^2}=\tan^{-1}\frac{2\text{x}}{1-\text{x}^2},$
$\Rightarrow\sin^{-1}\frac{2\tan\text{m}}{1+\tan^2\text{m}}-\cos^{-1}\frac{1-\tan^{2}\text{n}}{1+\tan^{2}\text{n}}=\tan^{-1}\frac{2\tan\text{y}}{1-\tan^2\text{y}}$
$\Rightarrow\sin^{-1}(\sin2\text{m})-\cos^{-1}(\cos2\text{n})=\tan^{-1}(\tan2\text{y})$
$\Big[\because\ \sin2\text{x}=\frac{2\tan\text{x}}{1+\tan^2\text{x}},\cos2\text{x}=\frac{1-\tan^2\text{x}}{1+\tan^2\text{x}}\Big]$
$\Rightarrow2\text{m}-2\text{n}=2\text{y}$
$\Rightarrow\text{m}-\text{n}=\text{y}$
$\Rightarrow\tan^{-1}\text{a}-\tan^{-1}\text{b}=\tan^{-1}\text{x}$
$[\because\ \text{a}=\tan\text{m},b=\tan\text{n}\text{ and }\text{x}=\tan\text{y}]$
$\Rightarrow\tan^{-1}\frac{\text{a}-\text{b}}{1+\text{ab}}=\tan^{-1}\text{x}$
$\Big[\because\ \tan^{-1}\text{x}-\tan^{-1}\text{y}=\tan^{-1}\frac{\text{x}-\text{y}}{1+\text{xy}}\Big]$
$\Rightarrow\frac{\text{a}-\text{b}}{1+\text{ab}}=\text{x}$
View full question & answer→Question 305 Marks
If $\cos^{-1}\frac{\text{x}}{2}+\cos^{-1}\frac{\text{y}}{3}=\alpha,$ then prove that $9\text{x}^2-12\text{xy}\cos\alpha+4\text{y}^2=36\sin^2\alpha.$
AnswerWe know
$\cos^{-1}\text{x}+\cos^{-1}\text{y}=\cos^{-1}\Big[\text{xy}-\sqrt{1-\text{x}^2}\sqrt{1-\text{y}^2}\Big]$
Now,
$\cos^{-1}\frac{\text{x}}{2}+\cos^{-1}\frac{\text{y}}{3}=\alpha$
$\Rightarrow\cos^{-1}\bigg[\frac{\text{x}}{2}\frac{\text{y}}{3}-\sqrt{1-\frac{\text{x}^2}{4}}\sqrt{1-\frac{\text{y}^2}{3}}\bigg]=\alpha$
$\Rightarrow\frac{\text{x}}{2}\frac{\text{y}}{3}-\sqrt{1-\frac{\text{x}^2}{4}}\sqrt{1-\frac{\text{y}^2}{3}}=\cos\alpha$
$\Rightarrow\text{xy}-\sqrt{4-\text{x}^2}\sqrt{9-\text{y}^2}=6\cos\alpha$
$\Rightarrow\sqrt{4-\text{x}^2}\sqrt{9-\text{y}^2}=\text{xy}-6\cos\alpha$
$\Rightarrow\big(4-\text{x}^2\big)\big(9-\text{y}^2\big)=\text{x}^2\text{y}^2+36\cos^2\alpha-12\text{xy}\cos\alpha$ [Squaring both sides]
$\Rightarrow36-4\text{y}^2-9\text{x}^2+\text{x}^2\text{y}^2=\text{x}^2\text{y}^2+36\cos^2\alpha-12\text{xy}\cos\alpha$
$\Rightarrow36-4\text{y}^2-9\text{x}^2+=36\cos^2\alpha-12\text{xy}\cos\alpha$
$\Rightarrow9\text{x}^2-12\text{xy}\cos\alpha+4\text{y}^2=36-36\cos^2\alpha$
$\Rightarrow9\text{x}^2-12\text{xy}\cos\alpha+4\text{y}^2=36\sin^2\alpha$
View full question & answer→Question 315 Marks
If $\Big(\sin^{-1}\text{x}\Big)^2+\Big(\cos^{-1}\text{x}\Big)^2=\frac{175\pi^2}{36},$ find x.
Answer$\Big(\sin^{-1}\text{x}\Big)^2+\Big(\cos^{-1}\text{x}\Big)^2=\frac{175\pi^2}{36}$ $\Rightarrow\Big(\sin^{-1}\text{x}\Big)^2+\Big(\frac{\pi}{2}-\sin^{-1}\text{x}\Big)^2=\frac{175\pi^2}{36}$ Let $\sin^{-1}\text{x}=\text{y}$ $\therefore\ (\text{y})^2+\Big(\frac{\pi}{2}-\text{y}\Big)^2=\frac{17\pi^2}{36}$ $\Rightarrow\text{y}^2+\frac{\pi^2}{4}+\text{y}^2-2\times\frac{\pi}{2}\times\text{y}=\frac{17\pi^2}{36}$ $\Rightarrow2\text{y}^2-\pi\text{y}=\frac{2\pi^2}{9}$$\Rightarrow18\text{y}^2-9\pi\text{y}-2\pi^2=0$
$\Rightarrow18\text{y}^2-12\pi\text{y}+3\pi\text{y}-2\pi\text{x}^2=0$
$\Rightarrow6\text{y}(3\text{y}-2\pi)+\pi(3\text{y}+2\pi)=0$
$\Rightarrow(3\text{y}-2\pi)(6\text{y}+\pi)=0$
$\Rightarrow\text{y}=-\frac{\pi}{6}$
[Neglecting $\text{y}=\frac{2}{3}\pi$ as it is not satisfying the question]$\therefore\text{x}=\sin\text{y}=\sin\Big(-\frac{\pi}{6}\Big)=-\frac{1}{2}$
View full question & answer→Question 325 Marks
Write the value of $\tan^{-1}\text{x}+\tan^{-1}\Big(\frac{1}{\text{x}}\Big)$ x < 0.
Answer$\tan^{-1}\text{x}+\tan^{1}\text{y}=\tan^{-1}\Big(\frac{\text{x}+\text{y}}{1-\text{xy}}\Big)$
When $\text{x}<0,\frac{1}{\text{x}}<0,$ then both are negative.
Let x = -y, y>0
Then,
$\tan^{1}{\text{x}}+\tan^{-1}\frac{1}{\text{x}}=\tan^{-1}(-\text{y})+\tan^{-1}\Big(-\frac{1}{\text{y}}\Big)$
$=-\Big(\tan^{-1}\text{y}+\tan^{-1}\frac{1}{\text{y}}\Big)$
$=-\tan^{-1}\bigg(\frac{\text{y}+\frac{1}{\text{y}}}{1-\text{y}\frac{1}{\text{y}}}\bigg),\text{y}>0$
$=-\tan^{-1}\Big(\frac{\text{y}^2+1}{0}\Big)$
$=-\tan^{-1}(\infty)$
$=-\tan^{-1}\Big(\tan\frac{\pi}{2}\Big)$
$=-\frac{\pi}{2}$
$\therefore\ \tan^{-1}\text{x}+\tan^{-1}\frac{1}{\text{x}}=-\frac{\pi}{2},\text{x}<0$
View full question & answer→Question 335 Marks
Write the following in the simplest form:
$\tan^{-1}\sqrt{\frac{\text{a}-\text{x}}{\text{a}+\text{x}}},-\text{a}<\text{x}<\text{a}$
AnswerLet, $\text{x}=\text{a}\cos\theta$
Now,
$\tan^{-1}\sqrt{\frac{\text{a}-\text{x}}{\text{a}+\text{x}}}=\tan^{-1}\sqrt{\frac{\text{a}-\text{a}\cos\theta}{\text{a}+\text{a}\cos\theta}}$
$=\tan^{-1}\sqrt{\frac{1-\cos\theta}{1+\cos\theta}}$
$=\tan^{-1}\sqrt{\frac{2\sin^2\frac{\theta}{2}}{2\cos^2\frac{\theta}{2}}}$
$=\tan^{-1}\Big(\tan\frac{\theta}{2}\Big)$
$=\frac{\theta}{2}$
$=\frac{1}{2}\cos^{-1}\Big(\frac{\text{x}}{\text{a}}\Big)$
$\therefore\ \tan^{-1}\sqrt{\frac{\text{a}-\text{x}}{\text{a}+\text{x}}}=\frac{\cos^{-1}\big(\frac{\text{x}}{\text{a}}\big)}{2}$
View full question & answer→Question 345 Marks
Write the following in the simplest form:
$\tan^{-1}\Big\{\frac{\sqrt{1+\text{x}^2}-1}{\text{x}}\Big\},\text{x}\neq0$
Answer$\tan^{-1}\Big\{\frac{\sqrt{1+\text{x}^2}-1}{\text{x}}\Big\},\text{x}\neq0$
Let, $\text{x}=\tan\theta$
$=\tan^{-1}\Big\{\frac{\sqrt{1+\tan^2\theta}-1}{\tan\theta}\Big\}$
$=\tan^{-1}\Big\{\frac{\sec\theta-1}{\tan\theta}\Big\}$ $\big\{\text{Since},1+\tan^2\theta=\sec^2\theta\big\}$
$=\tan^{-1}\Big\{\frac{1-\cos\theta}{\sin\theta}\Big\}$ $\Big\{\text{Since,}\sec\theta=\frac{1}{\cos\theta},\tan\theta=\frac{\sin\theta}{\cos\theta}\Big\}$
$=\tan^{-1}\Bigg\{\frac{\frac{2\sin^2\theta}{2}}{\frac{2\sin\theta}{2}\frac{\cos\theta}{2}}\Bigg\}$
$\Big\{\text{Since},1-\cos\theta=\frac{2\sin^2\theta}{2},\sin\theta=\frac{2\sin\theta}{2}\frac{\cos\theta}{2}\Big\}$
$=\tan^{-1}\bigg\{\frac{\frac{\sin\theta}{2}}{\frac{\cos\theta}{2}}\bigg\}$
$=\tan^{-1}\Big(\frac{\tan\theta}{2}\Big)$ $\Big\{\text{Since},\frac{\sin\theta}{\cos\theta}=\tan\theta\Big\}$
$=\frac{\theta}{2}$
$=\frac{1}{2}\tan^{-1}\text{x}$ $\{\text{Since},\tan\theta=\text{x}\Rightarrow\theta=\tan^{-1}\text{x}\}$
Hence,
$\tan^{-1}\Big\{\frac{\sqrt{1+\text{x}^2}-1}{\text{x}}\Big\}=\frac{1}{2}\tan^{-1}\text{x}$
View full question & answer→Question 355 Marks
Prove the following results:
$2\tan^{-1}\frac{1}{5}+\tan^{-1}\frac{1}{8}=\tan^{-1}\frac{4}{7}$
Answer$\text{L.H.S}=2\tan^{-1}\frac{1}{5}+\tan^{-1}\frac{1}{8}$
$=\tan^{-1}\frac{2\times\frac{1}{5}}{1-\big(\frac{1}{5}\big)^2}+\tan^{-1}\frac{1}{8}$ $\Big[\because\ 2\tan^{-1}\text{x}=\tan^{-1}\frac{2\text{x}}{1-\text{x}^2}\Big]$
$=\tan^{-1}\Bigg(\frac{\frac{2}{5}}{\frac{24}{25}}\Bigg)+\tan^{-1}\Big(\frac{1}{8}\Big)$
$=\tan^{-1}\Big(\frac{5}{12}\Big)+\tan^{-1}\Big(\frac{1}{8}\Big)$
$=\tan^{-1}\Bigg(\frac{\frac{5}{12}+\frac{1}{8}}{1-\frac{5}{12}\times\frac{1}{8}}\Bigg)$ $\Big[\text{Since }\tan^{-1}\text{x}+\tan^{-1}\text{y}=\tan^{-1}\frac{\text{x}+\text{y}}{1-\text{xy}}\Big]$
$=\tan^{-1}\Bigg(\frac{\frac{10+3}{24}}{\frac{96-5}{96}}\Bigg)$
$=\tan^{-1}\Big(\frac{13}{24}\times\frac{96}{91}\Big)$
$=\tan^{-1}\Big(\frac{4}{7}\Big)$
$=\text{R.H.S}$
Hence, $2\tan^{-1}\Big(\frac{1}{5}\Big)+\tan^{-1}\Big(\frac{1}{8}\Big)=\tan^{-1}\Big(\frac{4}{7}\Big)$
View full question & answer→Question 365 Marks
Solve the following equation for x:
$\tan^{-1}\Big(\frac{\text{x}-2}{\text{x}-4}\Big)+\tan^{-1}\Big(\frac{\text{x}+2}{\text{x}+4}\Big)=\frac{\pi}{4}$
Answer$\tan^{-1}\Big(\frac{\text{x}-2}{\text{x}-4}\Big)+\tan^{-1}\Big(\frac{\text{x}+2}{\text{x}+4}\Big)=\frac{\pi}{4}$
We know
$\because\ \tan^{-1}\text{x}+\tan^{-1}\text{y}=\tan^{-1}\Big(\frac{\text{x}+\text{y}}{1-\text{xy}}\Big)$
$\Rightarrow\tan^{-1}\Bigg(\frac{\frac{\text{x}-2}{\text{x}-4}+\frac{\text{x}+2}{\text{x}+4}}{1-\frac{\text{x}-2}{\text{x}-4}\times\frac{\text{x}+2}{\text{x}+4}}\Bigg)=\frac{\pi}4{}$
$\Rightarrow\tan^{-1}\Bigg(\frac{\frac{\text{x}^2+2\text{x}-8+\text{x}^2-2\text{x}-8}{(\text{x}-4)(\text{x}+4)}}{\frac{\text{x}^2-16-\text{x}^2+4}{(\text{x}-4)(\text{x}+4)}}\Bigg)=\frac{\pi}{4}$
$\Rightarrow\frac{2\text{x}^2-16}{-12}=\tan\frac{\pi}{4}$
$\Rightarrow\frac{2\text{x}^2-16}{-12}=1$
$\Rightarrow2\text{x}^2-16=-12$
$\Rightarrow2\text{x}^2=4$
$\Rightarrow\text{x}^2=2$
$\Rightarrow\text{x}=\pm\sqrt2$
View full question & answer→Question 375 Marks
Solve the following equation for x:
$\tan^{-1}(\text{x}+2)+\tan^{-1}(\text{x}-2)=\tan^{-1}\Big(\frac{8}{79}\Big),\text{x}>0$
AnswerWe know
$\tan^{-1}\text{x}+\tan^{-1}\text{y}=\tan^{-1}\Big(\frac{\text{x}+\text{y}}{1-\text{xy}}\Big)$
$\therefore\ \tan^{-1}(\text{x}+2)+\tan^{-1}(\text{x}-2)=\tan^{-1}\Big(\frac{8}{79}\Big)$
$\Rightarrow\tan^{-1}\Big(\frac{\text{x}+2+\text{x}-2}{1-(\text{x}+2)\times(\text{x}-2)}\Big)=\tan^{-1}\frac{8}{79}$
$\Rightarrow\frac{2\text{x}}{1-\text{x}^2+4}=\frac{8}{79}$
$\Rightarrow\frac{\text{x}}{5-\text{x}^2}=\frac{4}{79}$
$\Rightarrow79\text{x}=20-4\text{x}^2$
$\Rightarrow4\text{x}^2+79\text{x}-20=0$
$\Rightarrow4\text{x}^2+80\text{x}-\text{x}-20=0$
$\Rightarrow(4\text{x}-1)(\text{x}+20)=0$
$\Rightarrow\text{x}=\frac{1}{4}\text{ or }-20$
$\therefore\ \text{x}=\frac{1}{4}\ [\because\ \text{x}>0]$
View full question & answer→Question 385 Marks
Find the value of $\tan^{-1}\Big(\frac{\text{x}}{\text{y}}\Big)-\tan^{-1}\Big(\frac{\text{x-y}}{\text{x+y}}\Big)$
AnswerWe know
$\tan^{-1}\text{x}-\tan^{-1}\text{y}=\tan^{-1}\Big(\frac{\text{x-y}}{\text{1+xy}},\text{xy}\Big)>-1$
Now,
$\tan^{-1}\Big(\frac{\text{x}}{\text{y}}\Big)-\tan^{-1}\Big(\frac{\text{x-y}}{\text{x+y}}\Big)$
$=\tan^{-1}\begin{Bmatrix}\frac{\frac{\text{x}}{\text{y}}-\frac{\text{x-y}}{\text{x+y}}}{1+\frac{\text{x}}{\text{y}}\Big(\frac{\text{x-y}}{\text{x+y}}\Big)}\end{Bmatrix}$
$=\tan^{-1}\begin{Bmatrix}\frac{\frac{\text{x}^2+\text{xy}-\text{xy}+\text{y}^2}{\text{y}(\text{x+y})}}{\frac{\text{x}^2+\text{xy}-\text{xy}+\text{y}^2}{\text{y}(\text{x+y})}}\end{Bmatrix}$
$=\tan^{-1}1$
$=\tan^{-1}\Big(\tan\frac{\pi}{4}\Big)$
$=\frac{\pi}{4}$
$\therefore\ \tan^{-1}\Big(\frac{\text{x}}{\text{y}}\Big)-\tan^{-1}\Big(\frac{\text{x-y}}{\text{x+y}}\Big)=\frac{\pi}{4}$
View full question & answer→Question 395 Marks
Prove the following results:
$\tan^{-1}\frac{1}{7}+2\tan^{-1}\frac{1}{3}=\frac{\pi}{4}$
Answer$\tan^{-1}\Big(\frac{1}{7}\Big)+2\tan^{-1}\Big(\frac{1}{3}\Big)=\frac{\pi}{4}$ $\text{L.H.S}=\tan^{-1}\Big(\frac{1}{7}\Big)+2\tan^{-1}\Big(\frac{1}{3}\Big)$ $=\tan^{-1}\Big(\frac{1}{7}\Big)=\tan^{-1}\Bigg(\frac{2\big(\frac{1}{3}\big)}{1-\big(\frac{1}{3}\big)^2}\Bigg)$ $\Big\{\text{Since }2\tan^{-1}\text{x}=\frac{1}{2}\tan^{-1}\Big(\frac{2\text{x}}{1-\text{x}^2}\Big)\Big\}$ $=\tan^{-1}\Big(\frac{1}{7}\Big)+\tan^{-1}\Big(\frac{2}{3}\times\frac{9}{8}\Big)$ $=\tan^{-1}\Big(\frac{1}{7}\Big)+=\tan^{-1}\Big(\frac{3}{4}\Big)$ $=\tan^{-1}\Bigg(\frac{\frac{1}{7}+\frac{3}{4}}{1-\frac{1}{7}\times\frac{3}{4}}\Bigg)$ $\Big\{\text{Since}\tan^{-1}\text{x}+\tan^{-1}\text{y}=\tan^{-1}\Big(\frac{\text{x}+\text{y}}{1-\text{xy}}\Big)\Big\}$ $=\tan^{-1}\Bigg(\frac{\frac{25}{20}}{\frac{25}{20}}\Bigg)$ $=\tan^{-1}(1)$ $=\frac{\pi}{4}$ $=\text{R.H.S}$Hence, proved
View full question & answer→Question 405 Marks
Evaluate:
$\cos\Big(\sin^{-1}\frac{3}{5}+\sin^{-1}\frac{5}{13}\Big)$
Answer$\cos\Big(\sin^{-1}\frac{3}{5}+\sin^{-1}\frac{5}{13}\Big)$
$=\cos\Bigg[\sin^{-1}\Bigg(\frac{3}{5}\sqrt{1-\Big(\frac{5}{13}\Big)^2}+\frac{5}{13}\sqrt{1-\Big(\frac{3}{5}\Big)^2}\Bigg)\Bigg]$
$\Big\{\text{Since }\sin^{-1}\text{x}+\sin^{-1}\text{y}=\sin^{-1}\Big[\text{x}\sqrt{1-\text{y}^2}+\text{y}\sqrt{1-\text{x}^2}\Big]\Big\}$
$=\cos\Big[\sin^{-1}\Big(\frac{3}{5}\times\frac{12}{13}+\frac{5}{13}\times\frac{4}{5}\Big)\Big]$
$=\cos\Big[\sin^{-1}\Big(\frac{56}{65}\Big)\Big]$
$=\cos\Bigg[\cos^{-1}\Bigg(\sqrt{1-\Big(\frac{56}{65}\Big)^2}\Bigg]$
$\Big\{\text{Since }\sin^{-1}\text{x}=\cos^{-1}\Big(\sqrt{1-\text{x}^2}\Big)\Big\}$
$=\cos\Big[\cos^{-1}\Big(\frac{33}{65}\Big)\Big]$
$=\frac{33}{65}$ $\Big\{\text{Since }\cos\big(\cos^{-1}\text{x}\big)=\text{x}\text{ as }\text{x}\in[0,1]\Big\}$
Hence,
$\cos\Big(\sin^{-1}\frac{3}{5}+\sin^{-1}\frac{5}{13}\Big)=\frac{33}{65}$
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