Question

prove that cot square theta minus cos square theta is equals to COT square theta cos square theta​

Answer 1

Answer:

ans is in attachment.

please the attachment

Answer 2

Given,

$LHS=\cot ^{2} \theta - \cos ^{2} \theta$,

$RHS=\cot ^{2} \theta \cos ^{2} \theta$.

To prove,

LHS = RHS.

Solution,

Here, it is given that,

$\cot ^{2} \theta - \cos ^{2} \theta=\cot ^{2} \theta \cos ^{2} \theta \hfill ...(1)$

We have to prove that the above equation (1) is true. This can be done as follows.

Here, we have,

$LHS=\cot ^{2} \theta - \cos ^{2} \theta$

$=\frac{\cos^{2} \theta }{\sin^{2} \theta} - \cos ^{2} \theta \hfill (\because \cot^2 \theta = \frac{\cos^{2} \theta }{\sin^{2} \theta} )$

$=\frac{\cos^{2} \theta -(\cos ^{2} \theta)( \sin^{2} \theta) }{\sin^{2} \theta}$

$=(\cos^{2} \theta )\frac{1 - \sin^{2} \theta }{\sin^{2} \theta}$

$=\cos^{2} \theta \frac{ \cos^{2} \theta}{\sin^{2} \theta} \hfill (\because 1-\sin^2 \theta =\cos^{2} \theta)$

$=\cot ^{2} \theta \cos ^{2} \theta$

$\implies LHS =\cot ^{2} \theta \cos ^{2} \theta \hfill ...(2)$

Further, since we have,

$RHS=\cot ^{2} \theta \cos ^{2} \theta \hfill ...(3)$

From eq. (2) and (3), we can see that,

$LHS =RHS$

that is,

$\cot ^{2} \theta- \cos ^{2} \theta=\cot ^{2} \theta \cos ^{2} \theta$.

Hence, it is proved that $\cot ^{2} \theta- \cos ^{2} \theta=\cot ^{2} \theta \cos ^{2} \theta$.

#SPJ2