Question

6. If sin-¹3/5=A then find cot A.

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Answer 1

$\Large\textrm{ \cot A=\pm\dfrac{4}{3} }$

$\Large\textrm{Explained.}$

$\rm{(f\circ f^{-1})(x)=x}$

$\rm{f(x)=\sin x}$

$\rm{f^{-1}(x)=\arcsin x}$

$\rm{\sin(\arcsin(\dfrac{3}{5}))=\sin A}$

$\rm{\therefore \dfrac{3}{5}=\sin A}$

We know,

$\cdots\longrightarrow\boxed{\rm{\sin^{2}\theta+\cos^{2}\theta=1}}$

Dividing both sides,

$\cdots\longrightarrow\boxed{\rm{\tan^{2}\theta+1=\dfrac{1}{\cos \theta}}}$

$\rm{\cos^{2}A=1-\left(\dfrac{3}{5}\right)^{2}}$

$\rm{\cos^{2}A=1-\dfrac{9}{25}}$

$\rm{\cos^{2}A=\dfrac{16}{25}}$

$\rm{\tan^{2}A+1=\dfrac{25}{16}}$

$\rm{\tan^{2}A=\dfrac{9}{16}}$

$\rm{\dfrac{1}{\tan^{2}A}=\dfrac{16}{9}}$

$\rm{\cot^{2}A=\dfrac{16}{9}}$

$\rm{\therefore \cot A=\pm\dfrac{4}{3}}$

The sign depends on which quadrant $\rm{A}$ lies on.