Question

Inverse circular functions,Principal values of ${sin}^{-1}x,{cos}^{-1}x,{tan}^{-1}x$.
${tan}^{-1}x+{tan}^{-1}y={tan}^{-1}\frac{x+y}{1-xy}$, $xy<1$
$\pi +{tan}^{-1}\frac{x+y}{1-xy}$, $xy>1$.
(a) $sin\left({cos}^{-1}\frac{3}{5}\right)$
(b) $cos\left({tan}^{-1}\frac{3}{4}\right)$ and $cos\left({tan}^{-1}x\right)$
(c) If $sin\left({cot}^{-1}\right(1+x\left)\right)=cos\left({tan}^{-1}x\right)$ then x is
(a) $1/2$
(b) $1$
(c) $0$
(d) $-1/2$
(d) $sin\left({cot}^{-1}x\right)$
(e) $sin\left(2{sin}^{-1}0.8\right)$

Answer 1

(a) $sin{cos}^{-1}\left(3/5\right)=sin{sin}^{-1}\sqrt{1-9/25}$
$=sin{sin}^{-1}\left(4/5\right)=4/5$.
[Here we have used the formula
${cos}^{-1}x={sin}^{-1}\sqrt{1-x^2}]$.
(b) We have ${tan}^{-1}x={cos}^{-1}\frac{1}{\surd (1+x^2)}$.........(1)
$\therefore$ $cos\left({tan}^{-1}\frac{3}{4}\right)=cos\left({cos}^{-1}\frac{1}{\surd (1+9/16)}\right)$
$=cos\left({cos}^{-1}\frac{4}{5}\right)=\frac{4}{5}$, by(1)
For the second part, we have
$cos{tan}^{-1}x=cos{cos}^{-1}\frac{1}{\surd (1+x^2)}=\frac{1}{\surd (1+x^2)}$
(c) Ans (d)
We know that ${cot}^{-1}t={sin}^{-1}\frac{1}{\sqrt{1+t^2}},t=x+1$
and ${tan}^{-1}p={cos}^{-1}\frac{1}{\sqrt{1+p^2}},p=x$
Hence from the given relation we get
$\frac{1}{\sqrt{1+{(x+1)}^2}}=\frac{1}{\sqrt{1+x^2}}$
or $x^2+2x+2=x^2+1$
or $2x=-1$ $\therefore$ $x=-\frac{1}{2}\Rightarrow$ (d)
(d) by $\S 3,{cot}^{-1}x={sin}^{-1}\frac{1}{\surd (1+x^2)}$
$sin\left({cot}^{-1}x\right)=\frac{1}{\surd (1+x^2)}$.
(e) $2{sin}^{-1}x={sin}^{-1}2x\sqrt{1-x^2}$.
$\therefore$ $sin\left(2{sin}^{-1}0.8\right)=2.\frac{8}{10}\sqrt{(1-\frac{64}{100})}=0$.