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) Find whether $x=2$ satisfies the equation

${tan}^{-1}\frac{x+1}{x-1}+{tan}^{-1}\frac{x-1}{x}={tan}^{-1}(-7)$

If not, then how should the equation be re-written ?

(b) ${tan}^{-1}\frac{4}{3}+{tan}^{-1}\frac{5}{6}+{tan}^{-1}\frac{39}{2}-\pi =.....$

(c) If $x_1,x_2,x_3,x_4$ are roots of equation $x^4-x^{3}sin2\beta +x^{2}cos2\beta -xcos\beta -sin\beta =0$, then prove that ${\sum }_{i=1}^4{tan}^{-1}{x}_1=\frac{\pi }{2}-\beta$

Answer 1

(a) ${tan}^{-1}\frac{\frac{x+1}{x-1}+\frac{x-1}{x}}{1-\frac{x+1}{x-1}.\frac{x-1}{x}}={tan}^{-1}(-7)$

$\therefore$ $\frac{2x^2-x+1}{1-x}=-7$ or $2x^2-x+1=-7+7x$

or $2x^2-8x+8=0$ or ${}^{}-4x+4=0$

or ${(x-2)}^2=0$, $\therefore$ $x=2$

But if we put $x=2$ in the given equation the L.H.S. is $+ive$ and R.H.S. is $-ive$. Hence $x=2$ does not satisfy. We will have to write the equation as

${tan}^{-1}\frac{x+1}{x-1}+{tan}^{-1}\frac{x-1}{x}=\pi +{tan}^{-1}(-7)$

Now $x=2$ will make both sides $+ive$

Note : Here $xy=\frac{x+1}{x-1}\frac{x-1}{x}=\frac{x+1}{x}>1$

$\therefore$ R.H.S. $=\pi +{tan}^{-1}(-7)$

(b) Since $ab>1$, combine first two terms

$\therefore$ L.H.S.$=\pi +{tan}^{-1}(-\frac{39}{2})+{tan}^{-1}\left(\frac{39}{2}\right)-\pi =0$

(c) $p_1=\sum x_1=sin2\beta$

$p_2=\sum x_1{x}_2=cos2\beta$

$p_3=\sum x_1{x}_2{x}_3=cos\beta$

$p_4=\sum x_1{x}_2{x}_3{x}_4=sin\beta$

${tan}^{-1}{x}_1+{tan}^{-1}{x}_2+{tan}^{-1}{x}_3+{tan}^{-1}{x}_4={tan}^{-1}\frac{p_1-p_3}{1-p_2+p_4}$......(1)

$p_1-p_3=sin2\beta -cos\beta =cos\beta (2sin\beta -1)$

$1-p_2+p_4=1-cos2\beta -sin\beta$

$=2{sin}^2\beta -sin\beta =sin\beta (2sin\beta -1)$

$\therefore$ ${tan}^{-1}\frac{p_1-p_3}{1-p_2+p_4}={tan}^{-1}\left(\frac{cos\beta }{sin\beta }\right)$

$={tan}^{-1}\left(cot\beta \right)={tan}^{-1}\left[tan(\frac{\pi }{2}-\beta )\right]=\frac{\pi }{2}-\beta$