Question 14 Marks
If $\text{T}_\text{n}=\sin^\text{n}\text{x}+\cos^\text{n}\text{x},$ Prove that $\frac{\text{T}_3-\text{T}_5}{\text{T}_1}=\frac{\text{T}_5-\text{T}_7}{\text{T}_3}$
AnswerWe Have $\text{T}_\text{n}=\sin^\text{n}\text{x}+\cos^\text{n}\text{x}\cdots(\text{i})$ To show: $\frac{\text{T}_3-\text{T}_5}{\text{T}_1}=\frac{\text{T}_5-\text{T}_7}{\text{T}_3}$
$\text{L.H.S}=\frac{\text{T}_3-\text{T}_5}{\text{T}_1}$
$=\frac{(\sin^3\text{x}+\cos^3\text{x})-(\sin^5\text{x}+\cos^5\text{x})}{\sin\text{x}+\cos\text{x}}$ [Substituting the value of $T_3, T_5$ and $T_1$_ From (i)] $=\frac{\sin^3\text{x}-\sin^5\text{x}+\cos^3\text{x}-\cos^5\text{x}}{\sin\text{x}+\cos\text{x}}$
$=\frac{\sin^3\text{x}-(1-\sin^2\text{x})+\cos^3\text{x}(1-\cos^2\text{x})}{\sin\text{x}+\cos\text{x}}$
$=\frac{\sin^3\text{x}\cos^2\text{x}+\cos^3\text{x}\sin^2\text{x}}{\sin\text{x}+\cos\text{x}}$
$\Big[\because1-\sin^2\text{x}=\cos^2\text{x}\text{ and }1-\cos^2\text{x}=\sin^2\text{x}\Big]$
$=\frac{\sin^2\text{x}\cos^2\text{x}+(\sin\text{x}+\cos\text{x})}{\sin\text{x}+\cos\text{x}}$
$=\sin^2\text{x}\cos^2\text{x}$
$\text{R.H.S}=\frac{\sin^5\text{x}+\cos^5\text{x}-(\sin^7\text{x}+\cos^7\text{x})}{\sin^3\text{x}+\cos^3\text{x}}$
$=\frac{\sin^5\text{x}-\sin^7\text{x}+\cos^5\text{x}-\cos^7\text{x}}{\sin^3\text{x}+\cos^3\text{x}}$
$=\frac{\sin^5\text{x}(1-\sin^2\text{x})+\cos^5\text{x}(1-\cos^2\text{x})}{\sin^3\text{x}+\cos^3\text{x}}$
$=\frac{\sin^5\text{x}\cos^2\text{x}+\cos^5\text{x}\sin^2\text{x}}{\sin^3\text{x}+\cos^3\text{x}}$
$=\frac{\sin^5\text{x}\cos^2\text{x}(\sin^2\text{x}+\cos^3\text{x})}{\sin^2\text{x}+\cos^2\text{x}}$
$=\sin^2\text{x}\cos^2\text{x}$
$\text{L.H.S=R.H.S }$
$\text{Proved}$
View full question & answer→Question 24 Marks
If $\text{T}_\text{n}=\sin^\text{n}\text{x}+\cos^\text{n}\text{x},$ Prove that $6\text{T}_{10}-15\text{T}_8+10\text{T}_6-1=0$
Answer$=-6\sin^2\text{x}\cos^2\text{x}(\sin^4\text{x}+\cos^4\text{x})+6\sin^4\cos^4\text{x}\\\ \ \ +9\sin^2\text{x}\cos^2\text{x}(\sin^4\text{x}+\cos^4\text{x})-3\sin^2\text{x}\cos^2\text{x}$ $=3\sin^2\text{x}\cos^2\text{x}(\sin^4\text{x}+\cos^4\text{x})=6\sin^4\text{x}\cos^4\text{x}-3\sin^2\text{x}\cos^2\text{x}$ $=3\sin^2\text{x}\cos^2\text{x}((\sin^2\text{x})^2+(\cos^2\text{x})^2\\\ \ +2\sin^2\text{x}\cos^2\text{x }2\sin^2\text{x}\cos^2)\\\ \ +6\sin^4\text{x}\cos^4\text{x}-3\sin^2\text{x}\cos^2\text{x}$ $($Adding and subtracting $2\sin^2\text{x}\cos^2\text{x})$$$ $=3\sin^2\text{x}\cos^2\text{x}((\sin^2\text{x}+\cos^2\text{x})^2-2\sin^2\text{x}\cos^2\text{x})\\\ \ +6\sin^4\text{x}\cos^4\text{x}-3\sin^2\text{x}\cos^2\text{x}$ $=3\sin^2\text{x}\cos^2\text{x}(1-2\sin^2\text{x}\cos^2\text{x})+6\sin^4\text{x}\cos^4\text{x}-3\sin^2\text{x}\cos^2\text{x}$ $=3\sin^2\text{x}\cos^2\text{x}-6\sin^4\text{x}\cos^4+6\sin^4\text{x}\cos^4\text{x}-3\sin^2\text{x}\cos^2\text{x}$ $=0$ $=\text{R.H.S}$ $\text{Proved}.$
View full question & answer→Question 34 Marks
If $\text{T}_\text{n}=\sin^\text{n}\text{x}+\cos^\text{n}\text{x},$ Prove that $2 \text{T}_6 - 3\text{ T}_4 + 1 = 0$
Answer$\text{L.H.S}=2\text{T}_6-3\text{T}_4+1$
$=2(\sin^6\text{x}+\cos^6\text{x})-3(\sin^4\text{x}+\cos^4\text{x})+1$
$=2((\sin^2\text{x})^3+(\cos^2\text{x})^3-2(\sin^2\text{x})^2+(\cos^2\text{x})^2)+1$
$=2((\sin^2\text{x}+\cos^2\text{x})(\sin^2\text{x})^2+(\cos^2\text{x})^2-(\sin^2\text{x}\cos^2\text{x}))$
$\ \ \ -3((\sin^2\text{x})^2+(\cos^2\text{x})^2+2\sin^2\text{x}\cos^2\text{x}-2\sin^2\text{x}\cos^2\text{x})=1$
$ [$Using $a^3 + b^3 =(a + b)(a^2 + b^2 - ab)$ and adding and subtracting $2\sin^2\text{x}\cos^2\text{x}\big]$
$=2((\sin^2\text{x}+\cos^2\text{x})^2-3\sin^2\text{x}\cos^2\text{x})-3(1-2\sin^2\text{x}\cos^2\text{x})+1$
$=2(1-3\sin^2\text{x}\cos^2\text{x})-3\sin^2\text{x}\cos^2\text{x}+1$
$=2-6\sin^2\text{x}\cos^2\text{x}-2+6\sin^2\text{x}\cos^2\text{x}$
$=0$
$=\text{R.H.S}$
$\text{Proved}.$
View full question & answer→Question 44 Marks
Prove that $\Bigg|\sqrt{\frac{1-\sin\text{x}}{1+\sin\text{x}}}+\sqrt{\frac{1+\sin\text{x}}{1-\sin\text{x}}}\Bigg|$ $=-\frac{2}{\cos\text{x}},$ where $\frac{\pi}{2}<\text{x}<\pi$
Answer$\text{L.H.S}=\Bigg|\sqrt{\frac{1-\sin\text{x}}{1+\sin\text{x}}}+\sqrt{\frac{1+\sin\text{x}}{1-\sin\text{x}}}\Bigg|$ $=\Bigg|\frac{(\sqrt{1-\sin\text{x}})^2+(1+\sin\text{x})^2}{\sqrt{(1+\sin\text{x})(1-\sin\text{x})}}\Bigg|$ $=\Bigg|\frac{1-\sin\text{x}+1+\sin\text{x}}{\sqrt{1-\sin^2\text{x}}}\Bigg|$ $=\Big|\frac{2}{\cos\text{x}}\Big|$ $\Big(\because1-\sin^2\text{x}=\cos^2\text{x}\Rightarrow\sqrt{1-\sin^2\text{x}=\cos\text{x}}\Big)$ $=\frac{-2}{\cos\text{x}}$ $\Big(\frac{\pi}{2}<\text{x}<\pi\Rightarrow\cos\text{x}<0\Big)$ $=\text{R.H.S}$
View full question & answer→Question 54 Marks
Prove the following identities: $\Big(\frac{1}{\sec^2\text{x}-\cos^\text{x}}+\frac{1}{\text{Cosec}^2\text{x}-\sin^2\text{x}}\Big)\sin^2\text{x}\cos^2\text{x}=\frac{1-\sin^2\text{x}\cos^2\text{x}}{2+\sin^2\text{x}\cos^2\text{x}}$
Answer$\text{L.H.S}=\Big(\frac{1}{\sec^2\text{x}-\cos^2\text{x}}+\frac{1}{\text{cosec}^2\text{x}-\sin^2\text{x}}\Big)\sin^2\text{x}\cos^2\text{x}$
$=\Bigg(\frac{1}{\frac{1}{\cos^2\text{x}}-\cos^2\text{x}}+\frac{1}{\sin^2\text{x}-\sin^2\text{x}}\Bigg)\sin^2\text{x}\cos^2\text{x}$
$\Bigg(\frac{1}{\frac{1-\cos^4\text{x}}{\cos^2\text{x}}}+\frac{1}{\frac{1-\sin^4\text{x}}{\sin^2\text{x}}}\Bigg)\sin^2\text{x}\cos^2\text{x}$
$=\Big(\frac{\cos^2\text{x}}{(1-\cos^2\text{x})(1+\cos^2\text{x})}+\frac{\sin^2\text{x}}{(1-\sin^2\text{x})(1+\sin^2\text{x})}\Big)\sin^2\text{x}\cos^2\text{x}$
$\left(\begin{array}{c}\text{Using}1-\text{a}^4=1-(\text{a}^2)^2\\ =(1-\text{a}^2)(1+\text{a}^2)\end{array}\right)$ $=\Big(\frac{\cos^2\text{x}}{\sin^2\text{x}(1+\cos^2\text{x})}+\frac{\sin^2\text{x}}{\cos^2\text{x}(1+\sin^2)}\Big)\sin^2\text{x}\cos^2\text{x}$
$\left(\begin{array}{c}\text{Using }1-\cos^2\text{x}=\sin^2\text{x}\\\&1 \sin^2\text{x}=\cos^2\text{x}\end{array}\right)$
$=\Big(\frac{\cos^4\text{x}(1+\sin^\text{x})+\sin^4\text{x}(1+\cos^2\text{x})}{\sin^2\text{x}\cos^2\text{x}(1+\cos^2\text{x})(1+\sin^2\text{x})}\Big)\sin^2\text{x}\cos^2\text{x}$ $=\frac{\cos^4\text{x}+\sin^2\text{x}\cos^4\text{x}+\sin^4\text{x}+\cos^2\text{x}\sin^4\text{x}}{(1+\cos^2\text{x})(1+\sin^2\text{x})}$
$=\frac{(\cos^2\text{x})^2(\sin^2\text{x})^2+2\cos^2\text{x}\sin^2\text{x}-2\cos^2\text{x}\sin^2\text{x}+\sin^2\text{x}\cos^4\text{x}+\cos^2\text{x}\sin^4\text{x}}{(1+\cos^2\text{x})(1+\sin^2\text{x})}$
$(\text{adding an subtracting }2\cos^2\text{x}\sin^4\text{x})$
$=\frac{(\cos^2+\sin^2\text{x})^2-2\cos^2\text{x}\sin^2\text{x}+\sin^2\text{x}\cos^2\text{x}(\cos^2\text{x}+\sin^2\text{x})}{1+\sin^2\text{x}+\cos^2\text{x}+\sin^2\text{x}\cos^2\text{x}}$
$=\frac{1^2-2\cos^2\text{x}\sin^2\text{x}+\sin^2\text{x}\cos^2\text{x}.1}{1+1+\sin^2\text{x}+\cos^2\text{x}}$ $=\frac{1-\sin^2\text{x}\cos^2\text{x}}{2+\sin^2\text{x}\cos^2\text{x}}$
$=\text{R.H.S} $ Proved
View full question & answer→Question 64 Marks
If $\cos\text{x}-\sin\text{x}=\text{a}^3, \sec\text{x}-\cos\text{x}=\text{b}^3,$ than proved that $a^2b^2 (a^2 + b^2) = 1$.
AnswerGiven: $\text{cosec }\text{x}-\sin\text{x}=\text{a}^3,\sec\text{x}-\cos\text{x}=\text{b}^2$ To show: $\text{a}^2\text{b}^2(\text{a}^2+\text{b}^2)=1$ Since, $\text{cosec }\text{x}-\sin\text{x}=\text{a}^3$ $\Rightarrow\frac{1}{\sin\text{x}}-\sin\text{x}=\text{a}^3$
$\Big(\because\text{cosec }\text{x}=\frac{1}{\sin\text{x}}\Big)$
$\Rightarrow\frac{1-\sin^2\text{x}}{\sin\text{x}}=\text{a}^3$
$\Rightarrow\frac{\cos^2\text{x}}{\sin\text{x}}=\text{a}^3$
$(\because1-\sin^2\text{x}=\cos^2\text{x})$
$\Rightarrow\text{a}=\frac{\cos\frac{2}{3}\text{x}}{\sin\frac{1}{3}\text{x}}$ Since, $\frac{1}{\cos\text{x}}-\cos\text{x}=\text{b}^3$ $\Big(\because\sec\text{x}=\frac{1}{\cos\text{x}}\Big)$
$\Rightarrow\frac{1-\cos^2\text{x}}{\cos\text{x}}=\text{b}^3$
$\Rightarrow\frac{\sin^2\text{x}}{\cos\text{x}}=\text{b}^3$
$(\because1-\cos^2\text{x}=\sin^2\text{x})$
$\Rightarrow\text{b}=\frac{\sin\frac{2}{3}\text{x}}{\cos\frac{1}{3}\text{x}}$ Now, $\text{a}^2\text{b}^2 \text{(a}^2 +\text{ b}^2)$
$=\frac{\cos\frac{4}{3}\text{x}}{\sin\frac{2}{3}\text{x}}\times\frac{\sin\frac{4}{3}\text{x}}{\cos\frac{2}{3}\text{x}}\Bigg(\frac{\cos\frac{4}{3}\text{x}}{\sin\frac{2}{3}\text{x}}+\frac{\sin\frac{4}{3}\text{x}}{\cos\frac{2}{3}\text{x}}\Bigg)$
$=\cos\frac{2}{3}\text{x}\times\sin\frac{2}{3}\text{x}\frac{\Big(\cos\frac{6}{3}\text{x}+\sin\frac{6}{3}\text{x}\Big)}{\sin\frac{2}{3}\text{x}.\cos\frac{2}{3}\text{x}}$ $=\cos^2\text{x}+\sin^2\text{x}$
$=1$ $\text{Proved}$
View full question & answer→Question 74 Marks
Prove the following identities: $\frac{\cos\text{x}}{1-\sin\text{x}}=\frac{1+\cos\text{x}+\sin\text{x}}{1+\cos\text{x}-\sin\text{x}}$
Answer$\text{R.H.S}=\frac{1+\cos\text{x}+\sin\text{x}}{1+\cos\text{x}-\sin\text{x}}$ $=\frac{\big(\big(1+\cos\text{x}\big)+\sin\text{x}\big)}{\big(1+\cos\text{x}\big)-\sin\text{x}}\times\frac{\big(\big(1+\cos\text{x}\big)+\sin\text{x}\big)}{\big(1+\cos\text{x}+\sin\text{x}\big)}$ $=\frac{\big(\big(1+\cos\text{x}\big)+\sin\text{x}\big)^{2}}{\big(1+\cos\text{x}\big)^{2}-\sin^{2}\text{x}}$$\Big(\text{using}\big(\text{a+b}\big)\big(\text{a+b}\big)=\big(\text{a+b}\big)^{2}\&\big(\text{a+b}\big)\big(\text{a-b}\big)=\text{a}^{2}\text{b}^{2}\Big)$ $=\frac{\big(1+\cos\text{x}\big)^{2}+\sin^{2}\text{x}+2\sin\text{x}\big(1+\cos\text{x}\big)}{1+\cos^{2}\text{x}+2\cos\text{x}-\sin^{2}\text{x}}$ $\big(\text{using}\big(\text{a+b}\big)^{2}=\text{a}^{2}+\text{b}^{2}+2\text{ab}\big)$ $=\frac{1+\cos^{2}\text{x}+2.1\cos\text{x}+\sin^{2}\text{x}+2\sin\text{x}\big(1+\cos\text{x}\big)}{1+\cos{2}\text{x}+2\cos\text{x}-\big(1-\cos^{2}\text{x}\big)}$$\big(\text{using}\sin^{2}\text{x}=1-\cos^{2}\text{x}\big)$ $=\frac{1+1+2\cos\text{x}+2\sin\text{x}\big(1+\cos\text{x}\big)}{1-1+\cos^{2}\text{x}+\cos^{2}\text{x}+2\cos\text{x}}$ $\big(\text{using}\sin^{2}\text{x}+\cos^{2}\text{x}\big)=1$ $=\frac{2+2\cos\text{x}+2\sin\text{x}\big(1+\cos\text{x}\big)}{2\cos^{2}\text{x}+2\cos\text{x}}$ $=\frac{2\big(1+\cos\text{x}\big)+2\sin\text{x}\big(1+\cos\text{x}\big)}{2\cos\text{x}\big(\cos\text{x}+1\big)}$ $=\frac{\big(1+\cos\text{x}\big)\big(2+2\sin\text{x}\big)}{2\cos\text{x}\big(1+\cos\text{x}\big)}$ $=\frac{1+\sin\text{x}}{\cos\text{x}}$ $=\frac{1+\sin\text{x}}{\cos\text{x}}\times\frac{1-\sin\text{x}}{1-\sin\text{x}}$ $\text{L.H.S}=\frac{\cos\text{x}}{1-\sin\text{x}}$
View full question & answer→Question 84 Marks
If $\text{a}=\frac{2\sin\text{x}}{1+\cos\text{x}+\sin\text{x}},$ then proved that $\frac{1-\cos\text{x}+\sin\text{x}}{1+\sin\text{x}}$ is also equal to a.
AnswerWe have, $\frac{2\sin\alpha}{1\cos\alpha+\sin\alpha}=\text{y}$ Now, $\frac{1-\cos\alpha+\sin\alpha}{1+\sin\alpha}=\frac{(1-\cos\alpha+\sin\alpha)}{(1+\sin\alpha)}.\frac{(1+\cos\alpha+\sin\alpha)}{(1+\cos\alpha+\sin\alpha)}$ $=\frac{(1+\sin\alpha)^2-\cos^2\alpha}{(1+\sin\alpha)(1+\sin\alpha+\cos\alpha)}$ $=\frac{1+\sin^2\alpha+2\sin\alpha-1+\sin^2\alpha}{(1+\sin\alpha)(1+\sin\alpha+\cos\alpha)}$ $=\frac{2\sin\alpha(1+\sin\alpha)}{(1+\sin\alpha)(1+\sin\alpha+\cos\alpha)}$ $=\frac{2\sin\alpha}{1+\sin\alpha+\cos\alpha}=\text{y}$
View full question & answer→Question 94 Marks
If $\text{a}=\sec\text{x}-\tan\text{x}$ and $\text{b}=\text{cosec x}+\cot\text{x},$ then show that $\text{ab}+\text{a} - \text{b}+ 1=0.$
Answer$\text{L.H.S}=\text{ab + a} - \text{b + }1$ $=(\sec\text{x}-\tan\text{x})(\text{cosec+}\cot)+\sec\text{x}-\tan\text{x}-\text{cosec }\text{x}-\cot\text{x}+1$ $=\Big(\frac{1}{\cos\text{x}}-\frac{\sin\text{x}}{\cos\text{x}}\Big)\Big(\frac{1}{\sin\text{x}}+\frac{\cos\text{x}}{\sin\text{x}}\Big)+\frac{1}{\cos\text{x}}-\frac{\sin\text{x}}{\cos\text{x}}-\frac{1}{\sin\text{x}}-\frac{\cos\text{x}}{\sin\text{x}}+1$ $=\frac{1}{\sin\text{x}\cos\text{x}}+\frac{1}{\cos\text{x}}\times\frac{\cos\text{x}}{\sin\text{x}}-\frac{\sin\text{x}}{\cos\text{x}}\\\ \ \ \times\frac{1}{\sin\text{x}}-\tan\text{x}\times\cot\text{x}+\frac{1}{\cos\text{x}}-\frac{\sin\text{x}}{\cos\text{x}}-\frac{1}{\sin\text{x}}$ $=\frac{1}{\sin\text{x}\cos\text{x}}+\frac{1}{\sin\text{x}}-\frac{1}{\cos\text{x}}-1+\frac{1}{\cos\text{x}}-\frac{\sin\text{x}}{\cos\text{x}}-\frac{1}{\sin\text{x}}-\frac{\cos\text{x}}{\sin\text{x}}+1$ $=\frac{1}{\sin\text{x}\cos\text{x}}-\frac{\sin\text{x}}{\cos\text{x}}-\frac{\cos\text{x}}{\sin\text{x}}$ $=\frac{1\sin^2\text{x}-\cos^2\text{x}}{\sin\text{x}.\cos\text{x}}$ $=\frac{1-(\cos^2\text{x}+\sin^2\text{x})}{\sin\text{x}.\cos\text{x}}$ $=\frac{1-1}{\sin\text{x}.\cos\text{x}}=0$ $=\text{R.H.S. Hence Proved}$
View full question & answer→Question 104 Marks
Prove the following identities: $\frac{1−\sin\text{x}\cos\text{x}}{\cos\text{x}(\sec\text{x}−\text{cosec}\text{x})}\cdot\frac{\sin^2\text{x}−\cos^2\text{x}}{\sin^3\text{x}\cos^3\text{x}}=\sin\text{x}$
Answer$\text{L.H.S}=\frac{1-\sin\text{x}\cos\text{x}}{\cos\text{x}\big(\sec\text{x}-\text{cosec}\text{x}\big)}.\frac{\sin^{2}\text{x}-\cos^{2}\text{x}}{\sin^{3}\text{x}+\cos^{3}\text{x}}$ $=\frac{1-\sin\text{x}\cos\text{x}}{\cos\text{x}\big(\frac{1}{\cos\text{x}}-\frac{1}{\sin\text{x}}\big)}\cdot\frac{\big(\sin\text{x}+\cos\text{x}\big)\big(\sin\text{x}-\cos\text{x}\big)}{\big(\sin\text{x}+\cos\text{x}\big)\big(\sin^{2}\text{x}+\cos^{2}\text{x}-\sin\text{x}\cos{\text{x}}\big)}$$$ $\begin{bmatrix}\text{Using a}^2-\text{b}^2=(\text{a - b)(a + b)} \\\text{and a}^3+\text{b}^3\text{(a}^2+\text{b}^2-\text{ab}) \end{bmatrix}$ $=\frac{\big(1-\sin\text{x}\cos\text{x}\big)}{\cos\text{x}\big(\frac{\sin\text{x}-\cos\text{x}}{\cos\text{x}\sin\text{x}}\big)}.\frac{\sin\text{x}-\cos\text{x}}{1-\sin\text{x}\cos\text{x}}$ $ \big(\because\sin^{2}\text{x}+\cos^{2}\text{x}=1\big)$ $=\frac{\cos\text{x}\sin\text{x}}{\cos\text{x}}$ $=\sin\text{x}$ $=\text{R.H.S}$ $\text{Proved}$
View full question & answer→Question 114 Marks
Prove the following identities: $\frac{2\sin\text{x}\cos\text{x}-\cos\text{x}}{1-\sin\text{x}+\sin^2\text{x}-\cos^2\text{x}}=\cot\text{x}$
Answer$\text{L.H.S}=\frac{2\sin\text{x}\cos\text{x}-\cos\text{x}}{1-\sin+\sin^2\cos^2\text{x}}$ $=\frac{\cos\text{x}(2\sin\text{x}-1)}{1-\cos^2\text{x}+\sin^2\text{x}-\sin\text{x}}$ $=\frac{\cos\text{x}(2\sin\text{x}-1)}{\sin^2\text{x}+\sin^2\text{x}-\sin\text{x}}$ $(\because1-\cos^2\text{x}=\sin^2\text{x})$ $=\frac{\cos\text{x}(2\sin\text{x}-1)}{2\sin^2\text{x}-\sin\text{x}}$ $=\frac{\cos\text{x}(2\sin\text{x}-1)}{\sin\text{x}(2\sin\text{x}-1)}$ $=\frac{\cos\text{x}}{\sin\text{x}}$ $=\cot\text{x}$ $=\text{R.H.S}$ $\text{Proved}$
View full question & answer→Question 124 Marks
Prove the following identities $:\frac{(1+\cot\text{x}+\tan\text{x})(\sin\text{x}+\cos\text{x})}{\sec^3\text{x}-\text{cosec}^3\text{ x}}=\sin^2\text{x}\cos^2\text{x}$
Answer$\text{L.H.S}=\frac{(1+\cot\text{x}+\tan\text{x})(\sin\text{x}+\cos\text{x})}{\sec^3\text{x}-\text{cosec}^3\text{ x}}$ $=\frac{\Big(1+\frac{\cos\text{x}}{\sin\text{x}}+\frac{\sin\text{x}}{\cos\text{x}}\Big)}{\Big(\frac{1}{\cos^3\text{x}}-\frac{1}{\sin^3\text{x}}\Big)}$
$\left(\begin{array}{c}\because\cot\text{x}=\frac{\cos\text{x}}{\sin\text{x}},\tan\text{x}=\frac{\sin\text{x}}{\cos\text{x}}\\ \sec\text{x}=\frac{1}{\cos\text{x}},\text{cosec }\text{x}=\frac{1}{\sin\text{x}}\end{array}\right)$
$=\Bigg(\frac{1+\frac{\cos^2\text{x}+\sin^2\text{x}}{\sin\text{x}\cos\text{x}}}{\frac{\sin^3\text{x}-\cos^3\text{x}}{\cos^3\text{x}\sin^3\text{x}}}\Bigg)(\sin\text{x}-\cos\text{x})$
$=\frac{(\sin\text{x}\cos\text{x}+1)\sin^3\text{x}\cos^3\text{x}}{\sin\text{x}\cos\text{x}.(\sin^3\text{x}-\cos^3\text{x})}\cdot(\sin\text{x}-\cos\text{x})$
$(\because\sin^2\text{x}+\cos^2\text{x}=1)$
$=\frac{(1+\sin\text{x}\cos\text{x})\sin^2\text{x}\cos^2\text{x}.(\sin\text{x}-\cos\text{x})}{(\sin\text{x}-\cos\text{x})(\sin^3\text{x}+\cos^2\text{x}+\sin\text{x}\cos\text{x})}$
$(\because\text{a}^3-\text{b}^3=(\text{a-b})(\text{a}^2+\text{b}^2+\text{ab})$
$=\frac{(1+\sin\text{x}\cos\text{x}).\sin^2\text{x}\cos^2\text{x}}{1+\sin\text{x}\cos\text{x}}$
$=\sin^2\text{x}\cos^2\text{x}$ $=\text{R.H.S}$
$\text{Proved}$
View full question & answer→Question 134 Marks
Prove the following identities: $\cos\text{x}(\tan\text{x}+2)(2\tan\text{x}+1)=2\sec\text{x}+5\sin\text{x}$
Answer$\text{L.H.S}=\cos\text{x}(\tan\text{x}+2)(2\tan\text{x}+1)$ $=\cos\text{x}\Big(\frac{\sin\text{x}}{\cos\text{x}}+2\Big)\Big(\frac{2\sin\text{x}}{\cos\text{x}}=1\Big)\Big(\because\tan\text{x}=\frac{\sin\text{x}}{\cos\text{x}}\Big)$ $=\cos\frac{(\sin\text{x}+2\cos\text{x})(2\sin\text{x}+\cos\text{x})}{\cos\text{x}.\cos\text{x}}$ $=\frac{(2\sin^2\text{x}+\sin\text{x}\cos\text{x}+4\sin\text{x}\cos\text{x}+2\cos^2\text{x})}{\cos\text{x}}$ $=\frac{2(\sin^2\text{x}+\cos^2\text{x})+5\sin\text{x}\cos\text{x}}{\cos\text{x}}$ $=\frac{2+5\sin\text{x}\cos\text{x}}{\cos\text{x}}$ $(\because\sin^2\text{x}+\cos^2\text{x})=1$ $=\frac{2}{\cos\text{x}}+\frac{5\sin\text{x}\cos\text{x}}{\cos\text{x}}$ $=2\sec\text{x}+5\sin\text{x}$ $=\text{R.H.S}$ $\text{Proved}$
View full question & answer→Question 144 Marks
Prove the following identities: $(1+\tan\alpha\tan\beta)^2+(\tan\alpha-\tan\beta)^2=\sec^2\alpha\sec^2\beta$
Answer$\text{L.H.S}=(1+\tan\alpha\tan\beta)^2+(\tan\alpha-\tan\beta)^2$ $=1(\tan\alpha+\tan\beta)^2+2.1\tan\alpha\tan\beta\\\ \ +(\tan\alpha)^2+(\tan\beta)^2-1\tan\alpha.\tan\beta$ $(\text{Using(a+b)}^2=\text{a}^2+\text{b}^2+2\text{ab and (a-b)}^2=\text{a}^2+\text{b}^2-2\text{ab})$ $=1+\tan^2\alpha+\tan^2\beta+2\tan\alpha\tan\beta+\tan^2\alpha+\tan^2\beta-2\tan\alpha\tan\beta$ $=1+\tan^2\alpha+\tan^2\alpha+\tan^2\beta+\tan^2\beta$ $=\sec^2\alpha+\tan^2\beta(1+\tan^2\alpha)$$(\because1+\tan^2\alpha=\sec^2\alpha)$ $=\sec^2\alpha+\tan^2\beta.\sec^2\alpha$ $=\sec^2\alpha(1+\tan^2\beta)$ $=\sec^2\alpha.\sec^2\beta$ $(\because1+\tan^2\beta=\sec^2\beta)$ $=\text{R.H.S}$ $\text{Proved}$
View full question & answer→Question 154 Marks
Prove the following identities: $(\sec\text{x}\sec\text{x}+\tan\text{x}\tan\text{y})^2-(\sec\text{x}\tan\text{y}+\tan\text{x}\sec\text{y})^2=1$
Answer$\text{L.H.S}=\big(\sec\text{x}\sec\text{x}+\tan\text{x}\tan\text{y}\big)^{2}-\big(\sec\text{x}\tan\text{y}+\tan\text{x}\sec\text{y}\big)^{2}$ $=\big(\sec\text{x}\sec\text{y}\big)^{2}+\big(\tan\text{x}\tan\text{y}\big)^{2}+2\sec\text{x}\sec\text{y}\tan\text{x}\tan\text{y}\\\ \ \ -\Big(\big(\sec\text{x}\tan\text{y}\big)^{2}+\big(\tan\text{x}\sec\text{y}\big)^{2}+2\sec\text{x}\tan\text{y}\tan\text{x}\sec\text{y}\Big)$ $\big[\text{using}\big(\text{a+b}\big)^{2}=\text{a}^{2}+\text{b}^{2}+2\text{ab}\big]$ $=\sec^{2}\text{x}\sec^{2}\text{y}+\tan^{2}\text{x}\tan^{2}\text{y}+2\sec\text{x}\sec\text{y}\tan{\text{x}}\tan{\text{y}}\\\ \ \ -\sec^{2}\text{x}\tan^{2}\text{y}-\tan^{2}\text{x}\sec^{2}\text{y}-2\sec\text{x}\sec\text{y}\tan\text{x}\tan\text{y}$ $\big[\text{using}\big(\text{ab}\big)^{2}=\text{a}^{2}\text{b}^{2}\big]$ $=\sec^{2}\text{x}\sec^{2}\text{y}-\sec^{2}\text{x}\tan^{2}\text{y}+\tan^{2}\text{x}\tan^{2}{\text{y}}-\tan^{2}\text{x}\sec^{2}\text{y}$ $=\sec^{2}\text{x}\big(\sec^{2}\text{y}-\tan^{2}\text{y}\big)+\tan^{2}\text{x}\big(\tan^{2}\text{y}-\sec^{2}\text{y}\big)$ $= \sec^{2}\text{x}1-\tan^{2}\text{x}1$ $\bigg[\because\sec^{2}\theta=1+\tan^{2}\theta\Rightarrow\sec^{2}\theta-\tan^{2}\theta=1\bigg]$ $=1+\tan^{2}\text{x}-\tan^{2}\text{x}$ $=1$ $=\text{R.H.S}$ $\text{Proved}$
View full question & answer→Question 164 Marks
If $\sin\text{x}+\cos\text{x}=0$ and x lies in the fourth quadrant, find $\sin \text{x } $and $\cos\text{x}.$
AnswerWe have:$\sin\text{x} +\cos\text{x} = 0$
$\Rightarrow \sin \text{x} = - \cos\text{x}$
$\Rightarrow \frac{\sin\text{x}}{\cos\text{x}} = -1$$\Rightarrow \tan\text{x} = - 1$
Now, x is in the fourth quadrant. In the fourth quadrant, $\cos\text{x}$ and $\sec\text{x}$ are positive and all the other four T - ratios are$\therefore\ =\sqrt{ 1+\tan^2\text{x}} = \sqrt{1+(-1)^2}=\sqrt2$
$\cos\text{x} = \frac{1}{\sec\text{x}}=\frac{1}{\sqrt2}$
And, $\sin\text{x}= -\sqrt{1-\cos^2\text{x}} = \sqrt{1-\Big(\frac{1}{\sqrt2}\Big)^2} =\frac{-1}{\sqrt2}$$\therefore\sin\text{x}= \frac{-1}{\sqrt2}$ and $\cos\text{x}= \frac{1}{\sqrt2}$
View full question & answer→Question 174 Marks
Prove the following identities: $\frac{\tan^3\text{x}}{1+\tan^2\text{x}}+\frac{\cot^3\text{x}}{1+\cot^2\text{x}}=\frac{1-2\sin^2\text{x}\cos^2\text{x}}{\sin\text{x}\cos\text{x}}$
Answer$\text{L.H.S}=\frac{\tan^{3}\text{x}}{1+\tan^{2}\text{x}}+\frac{\cot^{3}\text{x}}{1+\cot^{2}\text{x}}$ $=\frac{\sin^{3}\text{x}}{\cos^{3}\text{x}\bigg(1+\frac{\sin^{2}\text{x}}{\cos^{2}\text{x}}\bigg)}+\frac{\cos^{3}\text{x}}{\sin^{3}\text{x}\bigg(1+\frac{\cos^{2}\text{x}}{\sin^{2}\text{x}}\bigg)}$$$ $\Big(\because\tan\text{x}=\frac{\sin\text{x}}{\cos\text{x}}\text{ and }\cot\text{x}=\frac{\cos\text{x}}{\sin\text{x}}\Big)$ $=\frac{\sin^{3}\text{x}\cos^{2}\text{x}}{\cos^{3}\text{x}\big(\cos^{2}\text{x}+\sin^{2}\text{x}\big)}+\frac{\cos^{3}\text{x}\sin^{2}\text{x}}{\sin^{3}\text{x}\big(\sin^{2}\text{x}+\cos^{2}\text{x}\big)}$$$ $=\frac{\sin{3}\text{ x}}{\cos\text{x}}+\frac{\cos^{3}\theta}{\sin\text{x}}$ $\big(\because\cos^{2}\text{x}+\sin^{2}\text{x}=1\big)$ $=\frac{\sin^{4}\text{x}+\cos^{4}\text{x}}{\sin\text{x}\cos\text{x}}$ $=\frac{\big(\sin^{2}\text{x}\big)^{2}+\big(\cos^{2}\text{x}\big)^{2}+2\sin^{2}\text{x}\cos^{2}\text{x}-2\sin^{2}\text{x}\cos^{2}\text{x}}{\sin\text{x}\cos\text{x}}$ $\big(\text{adding and subtracting }2\sin^{2}\text{x}\cos^{2}\text{x}\big)$ $=\frac{\big(\sin^{2}\text{x}+\cos^{2}\text{x}\big)^{2}-2\sin^{2}\text{x}\cos^{2}\text{x}}{\sin\text{x}\cos\text{x}}$ $=\frac{1^{2}-2\sin^{2}\text{x}\cos^{2}\text{x}}{\sin\text{x}\cos\text{x}}$ $\big(\because\sin^{2}\text{x}+\cos^{2}\text{x}=1\big)$ $=\frac{1-2\sin^{2}\text{x}\cos^{2}\text{x}}{\sin\text{x}\cos\text{x}}$ $=\text{R.H.S}$ $\text{Proved}$
View full question & answer→Question 184 Marks
If $\sin\text{x}=\frac{12}{13}$ and x lies in the second quadrant, find the value of $\sec\text{x}+\tan\text{x}.$
AnswerWe have: $\sin\text{x}=\frac{12}{13}$ and x lie in the second quadrant.In the second quadrant, $\sin\text{x}$ and $\text{cosec}\text{ x}$ are positive and all the other four T ratios.$\therefore\cos\text{x}=-\sqrt{1-\sin^2\text{x}}\\$
$=-\sqrt{1-\Big(\frac{12}{13}\Big)^2}$
$=\frac{-5}{13}$
$\tan\text{x}=\frac{\sin\text{x}}{\cos\text{x}}$
$=\frac{\frac{12}{13}}{\frac{-5}{13}}$ $=\frac{-12}{5}$ And, $\sec\text{x}=\frac{1}{\cos\text{x}}$$=\frac{1}{\frac{-5}{13}}$
$=\frac{-13}{5}$
$\therefore\sec\text{x}+\tan\text{x}=\frac{-13}{5}+\frac{-12}{5}$
$ = - 5$
View full question & answer→Question 194 Marks
If $\sin\text{x}+\sin\cos\text{x}=\text{m},$ then prove that $\sin^6\text{x}+\cos^6\text{x}=\frac{4-3(\text{m}^2-1)^2}{4},$ where $\text{m}^2\leq2$
AnswerTo show: $\sin^6\text{x}+\cos^6\text{x}=\frac{4-3(\text{m}^2-1)^2}{4},$ $\text{ where m}^2\leq2$ Since, $\sin\text{x}+\cos\text{x}=\text{m}\cdots(\text{i})$ $\Rightarrow(\sin\text{x}+\cos\text{x})^2=\text{m}^2$ $\Rightarrow\sin^2\text{x}+\cos^2\text{x}+2\sin\text{x}\cos\text{x}=\text{m}^2$ $\Rightarrow1+2\sin\text{x}\cos\text{x}=\text{m}^2$ $(\because\sin^2\text{x}+\cos^2\text{x}=1)$ $\Rightarrow2\sin\text{x}\cos\text{x}=\text{m}^2-1$ $\Rightarrow\sin\text{x}\cos\text{x}=\frac{\text{m}^2-1}{2}\cdots(\text{ii})$ $\therefore\text{L.H.S}=\sin^6\text{x}+\cos^2\text{x}$ $=(\sin^2\text{x})^3+(\cos^3\text{x})^3$ $=(\sin^2\text{x}+\cos^2\text{x})(\sin^2\text{x})^2+(\cos^2\text{x})^2-\sin^2\text{x}\cos^2\text{x}$ $=1.((\sin^2\text{x})^2+(\cos^2\text{x})^2\\\ \ \ +2\sin^2\text{x}\cos^2\text{x}-2\sin^2\text{x}\cos^2\text{x}-\sin^2\text{x}\cos^2\text{x})$ $(\text{adding and subtracting }2\sin^2\text{x}\cos^2\text{x})$ $=(\sin^2\text{x}+\cos^2\text{x})^2-3\sin^2\text{x}\cos^2\text{x}$ $=1-3\sin^2\text{x}\cos^2\text{x}$ $=1-3(\sin\text{x}\cos\text{x})^2$ $=1-3\frac{(\text{m}^2-1)^2}{4}$ $\text{(from (ii))}$ $=\frac{4-3(\text{m}^2-1)^2}{4}, \text{ where m}^2\leq2$ $=\text{R.H.S}$ $\text{Proved}$
View full question & answer→Question 204 Marks
Prove the following identities: $\text{cosec}\text{ x}(\sec\text{x}−1)−\cot\text{x}(1−\cos\text{x})=\tan\text{x}−\sin\text{x}$
Answer$\text{L.H.S}=\text{cosec }\text{x}\big(\sec\text{x}-1\big)-\cot\big(1-\cos\text{x}\big)$ $=\frac{1}{\sin\text{x}}\Big(\frac{1}{\cos\text{x}}-1\Big)-\frac{\cos\text{x}}{\sin\text{x}}\big(1-\cos\text{x}\big)$$ \Big[\because\text{cosec }\text{x}=\frac{1}{\sin\text{x}},\sec\text{x}=\frac{1}{\cos\text{x}}\cot\text{x}=\frac{\cos\text{x}}{\sin\text{x}}\Big]$ $=\frac{\big(1-\cos\text{x}\big)}{\sin\text{x}\cos\text{x}}-\frac{\cos\text{x}\big(1-\cos\text{x}\big)}{\sin\text{x}}$ $=\frac{\big(1-\cos\text{x}\big)-\cos^{2}\text{x}\big(1-\cos\text{x}\big)}{\sin\text{x}\cos\text{x}}$ $=\frac{\big(1-\cos\text{x}\big)\big(1-\cos^{2}\text{x}\big)}{\sin\text{x}\cos\text{x}}$ $=\frac{\big(1-\cos\text{x}\big)\sin^{2}\text{x}}{\sin\text{x}\cos\text{x}}$ $=\big(1-\cos\text{x}\big)\frac{\sin\text{x}}{\cos\text{x}}$ $=\frac{\sin\text{x}}{\cos\text{x}}-\sin\text{x}$ $=\tan\text{x}-\sin\text{x}$ $\big(\because1-\cos^{2}\text{x}=\sin^{2}\text{x}\big)$ $=\text{R.H.S} $ $\text{proved}$
View full question & answer→Question 214 Marks
Prove the following identities $:\frac{\sin^3+\cos^3\text{x}}{\sin\text{x}+\cos\text{x}}+\frac{\sin^3\text{x}-\cos^3\text{x}}{\sin\text{x}-\cos\text{x}}=2$
Answer$\text{L.H.S}=\frac{\sin^{3}\text{x}+\cos^{3}\text{x}}{\sin\text{x}+\cos\text{x}}+\frac{\sin^{3}\text{x}-\cos^{3}\text{x}}{\sin\text{x}-\cos\text{x}}$
$=\frac{\big(\sin\text{x}+\cos\text{x}\big)\big(\sin^{2}\text{x}+\cos^{2}\text{x}-\sin\text{x}\cos\text{x}\big)}{\big(\sin\text{x}+\cos\text{x}\big)} +\frac{\big(\sin\text{x} \cos\text{x}\big)\big(\sin^{2}\text{x}+\cos^{2}\text{x}+\sin\text{x}\cos\text{x}\big)}{\big(\sin\text{x}-\cos\text{x}\big)}$
$\left(\begin{array}{c}\text{using }\text{a}^{3}+\text{b}^{3}=\big(\text{a+b}\big)\big(\text{a}^{2}+\text{b}^{2}-\text{ab}\big)\\\text{ and }\text{ a}^{3}-\text{b}^{3}=\big(\text{a - b}\big)\big(\text{a}^{2}+\text{b}^{2}+\text{ab}\big)\end{array}\right)$
$=\big(1-\sin\text{x}\cos\text{x}\big)+\big(1+\sin\text{x}\cos\text{x}\big)$
$\big(\because \sin^{2}\text{x}+\cos^{2}\text{x}=1\big)$
$=2$
$=\text{R.H.S}$
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