Question 15 Marks
Two cards are drawn simultaneously (or successively without replacement) from a well shuffled pack of 52 cards. Find the mean and variance of the number of kings.
AnswerLet X denote the number of kings obtained in two draws. X takes the values 0, 1, 2.
Let p denote probability of first king.
$\therefore \ \text{p}=\frac{4}{52},\text{q}=\frac{48}{52}$
$\text{P}(\text{X}=0)=\frac{48}{52}\times\frac{47}{51}=\frac{12}{13}\times\frac{47}{51}=\frac{188}{221}$
$\text{P}(\text{X}=1)=\frac{4}{52}\times\frac{48}{51}+\frac{48}{52}\times\frac{4}{51}$
$=\frac{16}{221}+\frac{16}{221}=\frac{32}{221}$
$\text{P}(\text{X}=2)=\frac{4}{52}\times\frac{3}{51}$
$=\frac{1}{13}\times\frac{1}{17}$
$=\frac{1}{221}$
$\therefore$ probability distribution is:
| X |
0 |
1 |
2 |
| P(X) |
$\frac{188}{221}$ |
$\frac{32}{221}$ |
$\frac{1}{221}$ |
Now,
Mean of $\text{X}=\text{E}(\text{X})=\sum\limits^{\text{n}}_{\text{i}=1}\text{x}_{\text{i}}{\text{ p(x}_{\text{i}}})$
$=0\times\frac{188}{221}+1\times\frac{32}{221}+2\times\frac{1}{221}$
$0+\frac{32}{221}+\frac{2}{221}=\frac{34}{221}$
Also, $\text{E}(\text{X})^2=\sum\limits^{\text{n}}_{\text{i}=1}\text{x}_{\text{i}}^2{\text{ p(x}_{\text{i}}})$
$=0^2\times\frac{188}{121}+1^2\times\frac{32}{221}+2^2\times\frac{1}{221}$
$=0+\frac{32}{221}+\frac{4}{221}=\frac{36}{221}$
$\therefore \ \text{var}(\text{X})=\text{E}(\text{X})^2-[\text{E}(\text{X})]^2$
$=\frac{36}{221}-\Big(\frac{34}{221}\Big)^2=\frac{36}{221}-\frac{1156}{48841}$
$=\frac{7956-1156}{48841}$
$=\frac{6800}{48841}$
$\therefore\sigma^2=\sqrt{\text{var}(\text{X})}=\frac{\sqrt{6800}}{221}=\frac{82.46}{221}=0.37$ View full question & answer→Question 25 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 35 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 45 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 55 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 65 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 75 Marks
Prove that:
$\sin^{-1}\frac{8}{17}+\sin^{-1}\frac{3}{5}=\tan^{-1}\frac{77}{36}$
Answer$\text{Let}\sin^{-1}\frac{8}{17}=x.\text{Then}, \sin x=\frac{8}{17}$
$\Rightarrow\cos x=\sqrt{1-\left(\frac{8}{17}\right)^2}=\sqrt{\frac{225}{289}}=\frac{15}{17}.$
$\therefore\tan x=\frac{8}{15}\Rightarrow x=\tan^{-1}\frac{8}{15}$
$\therefore\sin^{-1}\frac{8}{17}=\tan^{-1}\frac{8}{15}\ \dots\dots(1)$
$\text{Now}, \text{let}\sin^{-1}\frac{3}{5}=y.\text{Then},\ \sin y=\frac{3}{5}$
$\Rightarrow\cos y=\sqrt{1-\left(\frac{3}{5}\right)^2}=\sqrt{\frac{16}{25}}=\frac{4}{5}.$
$\therefore\tan y=\frac{3}{4}\Rightarrow y=\tan^{-1}\frac{3}{4}$
$\therefore\sin^{-1}\frac{3}{5}=\tan^{-1}\frac{3}{4} \dots\dots(2)$
Now, we have:
$\text{L.H.S.}=\sin^{-1}\frac{8}{17}+\sin^{-1}\frac{3}{5}$
$=\tan^{-1}\frac{8}{15}+\tan^{-1}\frac{3}{4}$ [Using (1) and (2)]
$=\tan^{-1}\frac{\frac{8}{15}+\frac{3}{4}}{1-\frac{8}{15}\times\frac{3}{4}}$
$=\tan^{-1}\left(\frac{32+45}{60-24}\right)$ $\bigg[\tan^{-1}x+\tan^{-1}y=\tan^{-1}\frac{x+y}{1-xy}\bigg]$
$=\tan^{-1}\frac{77}{36}=\text{R.H.S.}$
View full question & answer→Question 85 Marks
Prove that $\tan^{-1}\bigg(\frac{\sqrt{1+\text{x}^2}+\sqrt{1-\text{x}^2}}{\sqrt{1+\text{x}^2}-\sqrt{1-\text{x}^2}}\bigg)=\frac{\pi}{4}+\frac{1}{2}\cos^{-1}\text{x}^2.$
AnswerWe have, $\tan^{-1}\bigg(\frac{\sqrt{1+\text{x}^2}+\sqrt{1-\text{x}^2}}{\sqrt{1+\text{x}^2}-\sqrt{1-\text{x}^2}}\bigg)$ Let $\text{x}^2=\cos2\theta\Rightarrow\ \theta=\frac{1}{2}\cos^{-1}\text{x}^2$ $\therefore$ L.H.S. $=\tan^{-1}\Big[\frac{\sqrt{1+\cos2\theta}+\sqrt{1-\cos2\theta}}{\sqrt{1+\cos2\theta}-\sqrt{1-\cos2\theta}}\Big]$ $=\tan^{-1}\Big[\frac{\sqrt{2\cos^2\theta}+\sqrt{2\sin^2\theta}}{\sqrt{2\cos^2\theta}-\sqrt{2\sin^2\theta}}\Big]$ $=\tan^{-1}\Big[\frac{\sqrt{2}\cos\theta+\sqrt{2}\sin\theta}{\sqrt{2}\cos\theta-\sqrt{2}\sin\theta}\Big]$ $=\text{tan}^{-1}\Bigg(\frac{\frac{cos\theta}{cos\theta}+\frac{sin\theta}{cos\theta}}{\frac{cos\theta}{cos\theta}-\frac{sin\theta}{cos\theta}}\Bigg)$ $=\tan^{-1}\Big[\frac{1+\tan\theta}{1-\tan\theta}\Big]$$=\tan^{-1}\bigg[\frac{\tan\frac{\pi}{4}+\tan\theta}{1-\tan\frac{\pi}{4}\tan\theta}\bigg]$
$=\tan^{-1}\Big(\tan\Big(\frac{\pi}{4}+\theta\Big)\Big)$
$\Big[\because\theta=\frac{1}{2}\cos^{-1}\text{x}^2\big]$$=\frac{\pi}{4}+\theta=\frac{\pi}{4}+\frac{1}{2}\cos^{-1}\text{x}^2$
View full question & answer→Question 95 Marks
Prove that:
$\cos^{-1}\frac{4}{5}+\cos^{-1}\frac{12}{13}=\cos^{-1}\frac{33}{65}$
Answer$\text{Let}\ \cos^{-1}\frac{4}{5}=x.\text{Then}, \cos x=\frac{4}{5}$
$\Rightarrow\sin x=\sqrt{1-\bigg(\frac{4}{5}\bigg)^2}=\frac{3}{5}.$
$\therefore\tan x=\frac{3}{4}\Rightarrow x=\tan^{-1}\frac{3}{4}$
$\therefore \cos^{-1}\frac{4}{5}=\tan^{-1}\frac{3}{4} \dots\dots(1)$
Now, let $\cos^{-1}\frac{12}{13}=y.$ Then, $\cos y=\frac{12}{13}\Rightarrow\sin y=\frac{5}{13}.$
$\therefore\tan y=\frac{5}{12}\Rightarrow y=\tan^{-1}\frac{5}{12}$
$\therefore\cos^{-1}\frac{12}{13}=\tan^{-1}\frac{5}{12} \dots\dots(2)$
Let $\cos^{-1}\frac{33}{65}=z.$ Then, $\cos z=\frac{33}{65}\Rightarrow\sin z=\frac{56}{65}.$
$\therefore\tan z=\frac{56}{33}\Rightarrow z=\tan^{-1}\frac{56}{33}$
$\therefore\cos^{-1}\frac{33}{65}=\tan^{-1}\frac{56}{33} \dots\dots(3)$
Now, we will prove that:
$\text{L.H.S.}=\cos^{-1}\frac{4}{5}+\cos^{-1}\frac{12}{13}$
$=\tan^{-1}\frac{3}{4}+\tan^{-1}\frac{5}{12}$ [Using(1) and (2)]
$=\tan^{-1}\frac{\frac{3}{4}+\frac{5}{12}}{1-\frac{3}{4}.\frac{5}{12}}$ $ \bigg[\tan^{-1}x+\tan^{-1}y=\tan^{-1}\frac{x+y}{1-xy}\bigg]$
$=\tan^{-1}\frac{36+20}{48-15}$
$=\tan^{-1}\frac{56}{33}$
$=\tan^{-1}\frac{56}{33}\ [\text{by}(3)]$
$=\cos^{-1}\frac{33}{65}\ \text{R.H.S.}$
View full question & answer→Question 105 Marks
Find the value of $4\tan^{-1}\frac{1}{5}-\tan^{-1}\frac{1}{239}.$
Answer$4\tan^{-1}\frac{1}{5}-\tan^{-1}\frac{1}{239}$
$=2\Big(2\tan^{-1}\frac{1}{5}\Big)-\tan^{-1}\frac{1}{239}$
$=2\tan^{-1}\frac{\frac{2}{5}}{1-\big(\frac{1}{5}\big)^2}-\tan^{-1}\frac{1}{239}$
$\Big(\because\ 2\tan^{-1}\text{x}=\tan^{-1}\frac{2\text{x}}{1-\text{x}^2}\Big)$
$=2\tan^{-1}\frac{\frac{2}{5}}{\frac{24}{25}}-\tan^{-1}\frac{1}{239}$
$=2\tan^{-1}\frac{5}{12}-\tan^{-1}\frac{1}{239}$
$=\tan^{-1}\frac{2.\frac{5}{12}}{1-\big(\frac{5}{12}\big)^2}-\tan^{-1}\frac{1}{239}$
$\Big(\because\ 2\tan^{-1}\text{x}=\tan^{-1}\frac{2\text{x}}{1-\text{x}^2}\Big)$
$=\tan^{-1}\frac{144\times5}{119\times6}-\tan^{-1}\frac{1}{239}$
$=\tan^{-1}\frac{120}{119}-\tan^{-1}\frac{1}{239}$
$=\tan^{-1}\frac{\frac{120}{119}-\frac{1}{239}}{1+\frac{120}{119}.\frac{1}{239}}$
$\Big(\because\ \tan^{-1}\text{x}-\tan^{-1}\text{y}=\tan^{-1}\frac{\text{x}-\text{y}}{1+\text{xy}}\Big)$
$=\tan^{-1}\frac{120\times239-119}{119\times239+120}$
$=\tan^{-1}\frac{28680-119}{28441+120}$
$=\tan^{-1}\frac{28561}{28561}=\tan^{-1}1=\frac{\pi}{4}$
View full question & answer→Question 115 Marks
Solve the following equation for x:
$\tan^{-1}\frac{1}{4}+2\tan^{-1}\frac{1}{5}+\tan^{-1}\frac{1}{6}+\tan^{-1}\frac{1}{\text{x}}=\frac{\pi}{4}$
Answer$\tan^{-1}\text{x}+\tan^{-1}\text{y}=\tan^{-1}\Big(\frac{\text{x}+\text{y}}{1-\text{xy}}\Big)$
$\therefore\ \tan^{-1}\frac{1}{4}+2\tan^{-1}\frac{1}{5}+\tan^{-1}\frac{1}{6}+\tan^{-1}\frac{1}{\text{x}}=\frac{\pi}{4}$
$\Rightarrow\tan^{-1}\frac{1}{4}+\tan^{-1}\frac{1}{5}+\tan^{-1}\frac{1}{5}+\tan^{-1}\frac{1}{6}+\tan^{-1}\frac{1}{\text{x}}=\frac{\pi}{4}$
$\Rightarrow\tan^{-1}\Bigg(\frac{\frac{1}{4}+\frac{1}{5}}{1-\frac{1}{4}\times\frac{1}{5}}\Bigg)+\tan^{-1}\Bigg(\frac{\frac{1}{5}+\frac{1}{6}}{1-\frac{1}{5}\times\frac{1}{6}}\Bigg)+\tan^{-1}\frac{1}{\text{x}}=\frac{\pi}{4}$
$\Rightarrow\tan^{-1}\Bigg(\frac{\frac{9}{20}}{\frac{19}{20}}\Bigg)+\tan^{-1}\Bigg(\frac{\frac{11}{30}}{\frac{29}{30}}\Bigg)+\tan^{-1}\frac{1}{\text{x}}=\frac{\pi}{4}$
$\Rightarrow\tan^{-1}\Big(\frac{9}{19}\Big)+\tan^{-1}\Big(\frac{11}{29}\Big)+\tan^{-1}\frac{1}{\text{x}}=\frac{\pi}{4}$
$\Rightarrow\tan^{-1}\Bigg(\frac{\frac{9}{19}+\frac{11}{29}}{1-\frac{11}{29}\times\frac{9}{19}}\Bigg)+\tan^{-1}\frac{1}{\text{x}}=\frac{\pi}{4}$
$\Rightarrow\Big(\frac{235}{226}\Big)+\tan^{-1}\frac{1}{\text{x}}=\frac{\pi}{4}$
$\Rightarrow\tan^{-1}\Bigg(\frac{\frac{235}{226}+\frac{1}{\text{x}}}{1-\frac{235}{226}\times\frac{1}{\text{x}}}\Bigg)=\frac{\pi}{4}$
$\Rightarrow\frac{235\text{x}+226}{226\text{x}-235}=\tan\frac{\pi}{4}$
$\Rightarrow\frac{235\text{x}+226}{226\text{x}-235}=1$
$\Rightarrow235\text{x}+226=226\text{x}-235$
$\Rightarrow9\text{x}=-461$
$\Rightarrow\text{x}=-\frac{461}{9}$
View full question & answer→Question 125 Marks
Show that $\tan\Big(\frac{1}{2}\sin^{-1}\frac{3}{4}\Big)=\frac{4-\sqrt{7}}{3}$ and justify why the other value $\frac{4+\sqrt{7}}{3}$ is ignored?
AnswerWe have, $\tan\Big(\frac{1}{2}\sin^{-1}\frac{3}{4}\Big)$
Let $\frac{1}{2}\sin^{-1}\frac{3}{4}=\theta$ $\Rightarrow\ \sin^{-1}\frac{3}{4}=2\theta\Rightarrow\ \sin2\theta=\frac{3}{4}$
$\Rightarrow\ \frac{2\tan\theta}{1+\tan^2\theta}=\frac{3}{4}$
$\Rightarrow\ 3\tan^2\theta-8$ and $\theta+3=0$
$\Rightarrow\ \tan\theta=\frac{8\pm\sqrt{64-36}}{6}$
$\Rightarrow\ \tan\theta=\frac{8\pm\sqrt{28}}{6}$
$=\frac{8\pm2\sqrt{7}}{6}=\frac{4\pm\sqrt{7}}{3}$
Now, $-\frac{\pi}{2}\leq\sin^{-1}\frac{3}{4}\leq\frac{\pi}{2}$
$\Rightarrow\ \frac{-\pi}{4}\leq\frac{1}{2}\sin^{-1}\frac{3}{4}\leq\frac{\pi}{2}$
$\therefore\ \tan\Big(\frac{-\pi}{4}\Big)\leq\tan\Big(\frac{1}{2}\Big(\sin^{-1}\frac{3}{4}\Big)\Big)\leq\tan\frac{\pi}{4}$
$\Rightarrow\ -1\leq\tan\Big(\frac{1}{2}\sin^{-1}\frac{3}{4}\Big)\leq1$
$\Rightarrow\ \tan\theta=\frac{4-\sqrt{7}}{3}$
$\bigg(\tan\theta=\frac{4+\sqrt{7}}{3}>1,\ \text{which is not possible}\bigg)$
View full question & answer→Question 135 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 145 Marks
Prove that:
$\cos^{-1}\frac{12}{13}=\sin^{-1}\frac{3}{5}=\sin^{-1}\frac{3}{5}=\sin^{-1}\frac{56}{65}$
Answer$\text{Let}\ \sin^{-1}\frac{3}{5}=x.\text{Then},\sin x=\frac{3}{5}$
$\Rightarrow\cos x=\sqrt{1-\bigg(\frac{3}{5}\bigg)^2}=\sqrt{\frac{16}{25}}=\frac{4}{5}.$
$\therefore\tan x=\frac{3}{4}\Rightarrow x=\tan^{-1}\frac{3}{4}$
$\therefore\sin^{-1}\frac{3}{5}=\tan^{-1}\frac{3}{4} \dots\dots(1)$
$\text{Now let}\cos^{-1}\frac{12}{13}=y,\cos y=\frac{12}{13}\Rightarrow\sin y=\frac{5}{13}.$
$\therefore\tan y=\frac{5}{12}\Rightarrow y=\tan^{-1}\frac{5}{12}$
$\therefore\cos^{-1}\frac{12}{13}=\tan^{-1}\frac{5}{12} \dots\dots(2)$
$\text{Let}\sin^{-1}\frac{56}{65}=z.\text{Then},\sin z=\frac{56}{65}\Rightarrow\cos z=\frac {33}{65}.$
$\therefore\tan z=\frac{56}{33}\Rightarrow z=\tan^{-1}\frac{56}{33}$
$\therefore\sin^{-1}\frac{56}{65}=\tan^{-1}\frac{56}{33} \dots\dots(3)$
Now, we have:
$\text{L.H.S.}=\cos^{-1}\frac{12}{13}+\sin^{-1}\frac{3}{5}$
$=\tan^{-1}\frac{5}{12}+\tan^{-1}\frac{3}{4}$ [Using (1) and (2)]
$=\tan^{-1}\frac{\frac{5}{12}+\frac{3}{4}}{1-\frac{5}{12}.\frac{3}{4}}$ $ \bigg[\tan^{-1}x+\tan^{-1}y=\tan^{-1}\frac{x+y}{1-xy}\bigg]$
$=\tan^{-1}\frac{20+36}{48-15}$
$=\tan^{-1}\frac{56}{33}$
$=\sin^{-1}\frac{56}{65}=\text{R.H.S.}$ [Using (3)]
View full question & answer→Question 155 Marks
Prove that:
$\tan^{-1}\frac{63}{16}=\sin^{-1}\frac{5}{13}+\cos^{-1}\frac{3}{5}$
Answer$\text{Let}\sin^{-1}\frac{5}{13}=x.\text{Then},\sin x=\frac{5}{13}\Rightarrow\cos x=\frac{12}{13}.$
$\therefore \tan x=\frac{5}{12}\Rightarrow x=\tan^{-1}\frac{5}{12}$
$\therefore\sin^{-1}\frac{5}{13}=\tan^{-1}\frac{5}{12} \dots\dots(1)$
$\text{Let}\cos^{-1}\frac{3}{5}=y.\text{Then},\cos y=\frac{3}{5}\Rightarrow\sin y=\frac{4}{5}.$
$\therefore\tan y=\frac{4}{3}\Rightarrow y=\tan^{-1}\frac{4}{5}.$
$\therefore\cos^{-1}\frac{3}{5}=\tan^{-1}\frac{4}{3} \dots\dots(2)$
Using (1) and (2), we have
$\text{R.H.S.}=\sin^{-1}\frac{5}{13}+\cos^{-1}\frac{3}{5}$
$=\tan^{-1}\frac{5}{12}+\tan^{-1}\frac{4}{3}$
$=\tan^{-1}\Bigg(\frac{\frac{5}{12}+\frac{4}{3}}{1-\frac{5}{12}\times\frac{4}{3}}\Bigg)$ $\left[\tan^{-1}x+\tan^{-1}y=\tan^{-1}\frac{x+y}{1-xy}\right]$
$=\tan^{-1}\left(\frac{15+48}{36-20}\right)$
$=\tan^{-1}\frac{63}{16}$
= L.H.S.
View full question & answer→Question 165 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 175 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 185 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 195 Marks
If $a_1, a_2, a_3, ..., a_n$ is an arithmetic progression with common difference d, then evaluate the following expression.
$\tan\bigg[\tan^{-1}\Big(\frac{\text{d}}{1+\text{a}_1\text{a}_2}\Big)+\tan^{-1}\Big(\frac{\text{d}}{1+\text{a}_2\text{a}_3}\Big)+\tan^{-1}\Big(\frac{\text{d}}{1+\text{a}_3\text{a}_4}\Big)+\ .....\ +\tan^{-1}\Big(\frac{\text{d}}{1+\text{a}_{\text{n}-1}\text{a}_\text{n}}\Big)\bigg]$
AnswerWe have, $a_1 = a, a_2 = a + d, a_3 = a + 2d ....$
And $d = a_2 - a_1 = a_3 - a_2 = a_4 - a_3 = ..... = a_n - a_{n-1}$
Given that, $\tan\bigg[\tan^{-1}\Big(\frac{\text{d}}{1+\text{a}_1\text{a}_2}\Big)+\tan^{-1}\Big(\frac{\text{d}}{1+\text{a}_2\text{a}_3}\Big)+\tan^{-1}\Big(\frac{\text{d}}{1+\text{a}_3\text{a}_4}\Big)+\ .....\ +\tan^{-1}\Big(\frac{\text{d}}{1+\text{a}_{\text{n}-1}\text{a}_\text{n}}\Big)\bigg]$
$=\tan\bigg[\tan^{-1}\frac{\text{a}_2-\text{a}_1}{1+\text{a}_2.\text{a}_1}+\tan^{-1}\frac{\text{a}_3-\text{a}_2}{1+\text{a}_3.\text{a}_2}+\ ....\ +\tan^{-1}\frac{\text{a}_\text{n}-\text{a}_{\text{n}-1}}{1+\text{a}_\text{n}.\text{a}_{\text{n}-1}}\bigg]$
$=\tan\bigg[\Big(\tan^{-1}\text{a}_2-\tan^{-1}\text{a}_1\Big)+\Big(\tan^{-1}\text{a}_3-\tan^{-1}\text{a}_2\Big)+\ ....\ +\Big(\tan^{-1}\text{a}_{\text{n}}-\tan^{-1}\text{a}_{\text{n}-1}\Big)\bigg]$
$=\tan\Big[\tan^{-1}\text{a}_\text{n}-\tan^{-1}\text{a}_1\Big]$
$=\tan\Big[\tan^{-1}\frac{\text{a}_\text{n}-\text{a}_1}{1+\text{a}_\text{n}.\text{a}_1}\Big]$
$\bigg[\because\ \tan^{-1}\text{x}-\tan^{-1}\text{y}=\tan^{-1}\Big(\frac{\text{x}-\text{y}}{1+\text{xy}}\Big)\bigg]$
$=\frac{\text{a}_\text{n}-\text{a}_1}{1+\text{a}_\text{n}.\text{a}_1}\ \Big[\because\ \tan\big(\tan^{-1}\text{x}\big)=\text{x}\Big]$
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