sec 2 theta - cos2 theta = sin 2theta ( sec2theta +1 )
Answers
Answer:
Proved that \sec^{2}\theta - \cos^{2}\theta = \sin^{2}\theta(\sec^{2}\theta + 1 )sec
2
θ−cos
2
θ=sin
2
θ(sec
2
θ+1)
We have to prove that \sec^{2}\theta - \cos^{2}\theta = \sin^{2}\theta(\sec^{2}\theta + 1 )sec
2
θ−cos
2
θ=sin
2
θ(sec
2
θ+1)
Now, left hand side
= \sec^{2}\theta - \cos^{2}\thetasec
2
θ−cos
2
θ
= 1 + \tan^{2}\theta + \sin^{2}\theta - 11+tan
2
θ+sin
2
θ−1
{Since, we know that \sec^{2}\theta - \tan^{2}\theta = 1sec
2
θ−tan
2
θ=1 and \sin^{2}\theta + \cos^{2}\theta = 1sin
2
θ+cos
2
θ=1 }
= \tan^{2}\theta + \sin^{2}\thetatan
2
θ+sin
2
θ
= \frac{\sin^{2}\theta }{\cos^{2}\theta} + \sin^{2}\theta
cos
2
θ
sin
2
θ
+sin
2
θ
= \sin^{2}\theta[\frac{1}{\cos^{2}\theta} + 1 ]sin
2
θ[
cos
2
θ
1
+1]
= \sin^{2}\theta(\sec^{2}\theta + 1 )sin
2
θ(sec
2
θ+1)
= right hand side