Question

Question 8 Prove that [ cos (π+x) . cos (-x) ] / [ sin (π - x) . cos (π/2 +x) ] = (cot (x))^2

Class X1 - Maths -Trigonometric Functions Page 73

Answer 1

LHS = {cos(π+x).cos(-x)}/{sin(π-x).cos(π/2-x)}
we know,
cos(π+x) = -cosx
sin(π-x) = sinx
cos(-x) = cosx
cos(π/2 + x) = -sinx, use this above .

LHS = {(-cosx).cosx}/{sinx.(-sinx}
= cos²x/sin²x
= cot²x = RHS

Answer 2

Answer and explanation:

To prove : $\frac{\cos(\pi +x)\csdot \cos(-x)}{\sin(\pi-x)\cdot \cos(\frac{\pi}{2}+x)}=(\cot(x))^2$

Proof :

$LHS=\frac{\cos(\pi +x)\csdot \cos(-x)}{\sin(\pi-x)\cdot \cos(\frac{\pi}{2}+x)}$

Applying trigonometric properties,

$\cos(\pi +x)=-\cos x$

$\sin(\pi-x)=\sin x$

$\cos(-x)=\cos x$

$\cos(\frac{\pi}{2}+x)=-\sin x$

Substitute all the values in the expression,

$LHS=\frac{-\cos x\csdot \cos x}{\sin x\cdot -\sin x}$

$LHS=\frac{-\cos^2 x}{-\sin^2 x}$

$LHS=(\frac{\cos x}{\sin x})^2$

$LHS=(\cot(x))^2$

$LHS=RHS$

Hence proved.