Question

Plz Solve this question

Answer 1

Answer:

I tried my best but I can't solve it sorry

Answer 2

Answer: u can choose any one which is easier to u

Step-by-step explanation:

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

                                                          OR

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