Question

Prove cotx=sinxsin(\frac{\pi}{2}-x)+cos^{2}xcotx [/tex]

Answer 1

We have to prove that :

$Cot(x)=Sin(x)Sin(\frac{\pi}{2}-x)+Cos^{2}(x)Cot(x)$

For this, let us take the Right Hand Side (RHS) and then prove the Left Hand Side (LHS).

RHS=$Sin(x)Sin(\frac{\pi}{2}-x)+Cos^{2}(x)Cot(x)$

=$Sin(x)Cos(x)+Cos^{2}(x)Cot(x)$ ($\because Sin(\frac{\pi}{2}-x)=Cos(x)$)

=$Sin(x)Cos(x)+Cos^{2}(x)\frac{Cos(x)}{Sin(x)}$ ($\because Cot(x)=\frac{Cos(x)}{Sin(x)}$)

=$\frac{Sin^2(x)Cos(x)+Cos^2(x)Cos(x)}{Sin(x)}$

Now, we know that $Cos(x)$ is common in the numerator and thus,

$\frac{Cos(x)(Sin^2(x)+Cos^2(x))}{Sin(x)}$

=$\frac{Cos(x)}{Sin(x)}$ ($\because Sin^2(x)+Cos^2(x)=1$)

=$Cot(x)$=LHS

Hence, LHS=RHS and thus the required equation has been proved.