Question

Sin square 28+ sin square 62 + tan square 38 - cot square 52 + 1/4 sec square 30

Answer 1

prefer the image for the solution

Answer 2

$\sin^2 28+\sin^2 62+\tan^2 38-\cot^2 52+\dfrac{1}{4} \sec^2 30=\dfrac{4}{3}$

Step-by-step explanation:

We have,

$\sin^2 28+\sin^2 62+\tan^2 38-\cot^2 52+\dfrac{1}{4} \sec^2 30$

To find, $\sin^2 28+\sin^2 62+\tan^2 38-\cot^2 52+\dfrac{1}{4} \sec^2 30=?$

∴ $\sin^2 28+\sin^2 62+\tan^2 38-\cot^2 52+\dfrac{1}{4} \sec^2 30$

$=\sin^2 28+\sin^2 (90-28)+\tan^2 38-\cot^2 (90-38)+\dfrac{1}{4} \times(\dfrac{2}{\sqrt{3}})^2$

[ ∵ $\sec 30=\dfrac{2}{\sqrt{3}}$]

$=\sin^2 28+\cos^2 28+\tan^2 38-\tan^2 38+\dfrac{1}{4} \times \dfrac{4}{3} )$

[ ∵ $\sin (90-A)=\cos A$ and $\cot (90-A)=\tan A$]

$=(\sin^2 28+\cos^2 28)+(\tan^2 38-\tan^2 38)+\dfrac{1}{4} \times \dfrac{4}{3} )$

$=1+0+\dfrac{1}{3}$

=$\dfrac{3+1}{3} =\dfrac{4}{3}$

Hence, $\sin^2 28+\sin^2 62+\tan^2 38-\cot^2 52+\dfrac{1}{4} \sec^2 30=\dfrac{4}{3}$