⇒Trigonometric Functions

Correct answer option is (C)

Note that ${\cot }^{-1}x={\sin }^{-1}\left(\frac{1}{\sqrt{1+x^2}}\right)$ and

$\begin{array}{rl}& {\tan }^{-1}(1+x)={\cos }^{-1}\left(\frac{1}{\sqrt{1+(1+x{)}^2}}\right)\\ & \therefore \sin ({\cot }^{-1}(x))=\cos ({\tan }^{-1}(1+x))\\ & \Rightarrow \sin ({\sin }^{-1}\left(\frac{1}{\sqrt{1+x^2}}\right))=\cos ({\cos }^{-1}\left(\frac{1}{\sqrt{1+(1+x{)}^2}}\right))\\ & \Rightarrow \frac{1}{\sqrt{1+x^2}}=\frac{1}{\sqrt{1+(1+x{)}^2}}\\ & \Rightarrow 1+(1+x{)}^2=1+x^2\\ & \Rightarrow x=\frac{-1}{2}\end{array}$

Correct answer option is (C)

$\begin{array}{rl}& {\tan }^{-1}\left(\frac{1-x}{1+x}\right)=\frac{1}{2}{\tan }^{-1}x\\ & \Rightarrow {\tan }^{-1}(1)-{\tan }^{-1}(x)=\frac{1}{2}{\tan }^{-1}x\\ & \Rightarrow \frac{\pi }{4}=\frac{3}{2}{\tan }^{-1}x\\ & \Rightarrow x=\tan \left(\frac{\pi }{6}\right)=\frac{1}{\sqrt{3}}\end{array}$

Correct answer option is (B)

$\begin{array}{rl}& x=\mathrm{cosec}({\tan }^{-1}(\cos ({\cot }^{-1}(\sec ({\sin }^{-1}\mathrm{a})))))\\ & =\mathrm{cosec}({\tan }^{-1}(\cos ({\cot }^{-1}(\sec ({\sec }^{-1}\frac{1}{\sqrt{1-{\mathrm{a}}^2}})))))\\ & =\mathrm{cosec}({\tan }^{-1}(\cos ({\cot }^{-1}\left(\frac{1}{\sqrt{1-{\mathrm{a}}^2}}\right))))\\ & =\mathrm{cosec}({\tan }^{-1}(\cos ({\cos }^{-1}\frac{1}{\sqrt{2-{\mathrm{a}}^2}})))\\ & =\mathrm{cosec}({\tan }^{-1}\left(\frac{1}{\sqrt{2-{\mathrm{a}}^2}}\right))\\ & =\mathrm{cosec}\left({\mathrm{cosec}}^{-1}\left(\sqrt{3-{\mathrm{a}}^2}\right)\right)\\ & \therefore x=\sqrt{3-{\mathrm{a}}^2}\\ & \therefore x^2+{\mathrm{a}}^2=3\end{array}$

Correct answer option is (A)

$\begin{array}{ll}& \sin ({\cot }^{-1}x)\\ & \text{ Let }{\cot }^{-1}x=t\\ \therefore & x={\cot }^t\\ \therefore & 1+{\cot }^{2}t=1+x^2\\ \therefore & {\mathrm{cosec}}^{2}t=1+x^2\\ \therefore & \mathrm{cosec}t=\sqrt{1+x^2}\\ \therefore & \sin t=\frac{1}{\sqrt{1+x^2}}\\ \therefore & t={\sin }^{-1}\left(\frac{1}{\sqrt{1+x^2}}\right)\\ \therefore & \sin ({\cot }^{-1}x)=\sin ({\sin }^{-1}\left(\frac{1}{\sqrt{1+x^2}}\right))\\ & =\frac{1}{\sqrt{1+x^2}}\end{array}$

Correct answer option is (A)

$\begin{array}{rl}& {\cos }^{-1}\sqrt{\mathrm{p}}-{\cos }^{-1}\sqrt{1-\mathrm{p}}+{\cos }^{-1}\sqrt{1-\mathrm{q}}=\frac{3\pi }{4}\\ & \text{ Let }\mathrm{t}={\cos }^{-1}\sqrt{\mathrm{p}}\\ & \Rightarrow \mathrm{p}={\cos }^2\mathrm{t}\\ & \Rightarrow \mathrm{p}=1-{\sin }^2\mathrm{t}\\ & \Rightarrow \sin \mathrm{t}=\sqrt{1-\mathrm{p}}\\ & \Rightarrow \mathrm{t}={\sin }^{-1}\sqrt{1-\mathrm{p}}\end{array}$

$\Rightarrow {\cos }^{-1}\sqrt{\mathrm{p}}={\sin }^{-1}\sqrt{1-\mathrm{p}}$

$\therefore$ Given equation becomes

$\begin{array}{ll}\therefore & {\sin }^{-1}\sqrt{1-p}-{\cos }^{-1}\sqrt{1-p}+{\cos }^{-1}\sqrt{1-q}=\frac{3\pi }{4}\\ \therefore & \frac{\pi }{2}+{\cos }^{-1}\sqrt{1-q}=\frac{3\pi }{4}\dots [\because {\cos }^{-1}a+{\sin }^{-1}a=\frac{\pi }{2}]\\ \therefore & {\cos }^{-1}\sqrt{1-q}=\frac{-\pi }{4}\\ \therefore & \sqrt{1-q}=\cos (-\frac{\pi }{4})\\ \therefore & q=1-\frac{1}{2}\\ \therefore & q=\frac{1}{2}\end{array}$