Question

prove the following 1 upon 1 + sin square theta + 1 upon 1 + cos square theta + 1 upon 1 + secant square theta + 1 upon 1 + cosec square theta is equal to 2

Answer 1

Step-by-step explanation:

Proved

Step-by-step explanation:

(\frac{1}{Sec^2\theta - Cos^2\theta} + \frac{1}{Cosec^2\theta - Sin^2\theta} )\times Sin^2\theta Cos^2\theta = \frac{1 -Sin^2\theta Cos^2\theta}{2 + Sin^2\theta Cos^2\theta}(Sec2θ−Cos2θ1+Cosec2θ−Sin2θ1)×Sin2θCos2θ=2+Sin2θCos2θ1−Sin2θCos2θ

LHS = (\frac{1}{Sec^2\theta - Cos^2\theta} + \frac{1}{Cosec^2\theta - Sin^2\theta}) \times Sin^2\theta Cos^2\theta(Sec2θ−Cos2θ1+Cosec2θ−Sin2θ1)×Sin2θCos2θ

Secθ = 1/Cosθ  & Cosecθ = 1/Sinθ

= (\frac{1}{\frac{1}{Cos^2\theta} - Cos^2\theta} + \frac{1}{\frac{1}{Sin^2\theta} - Sin^2\theta}) \times Sin^2\theta Cos^2\theta(Cos2θ1−Cos2θ1+Sin2θ1−Sin2θ1)×Sin2θCos2θ

= (\frac{Cos^2\theta}{1 - Cos^4\theta} + \frac{Sin^2\theta}{1 - Sin^4\theta}) \times Sin^2\theta Cos^2\theta(1−Cos4θCos2θ+1−Sin4θSin2θ)×Sin2θCos2θ

= (\frac{Cos^2\theta}{(1 + Cos^2\theta)(1 - Cos^2\theta)} + \frac{Sin^2\theta}{(1 + Sin^2\theta)(1 - Sin^2\theta)}) \times Sin^2\theta Cos^2\theta((1+Cos2θ)(1−Cos2θ)Cos2θ+(1+Sin2θ)(1−Sin2θ)Sin2θ)×Sin2θCos2θ

= (\frac{Cos^2\theta}{(1 + Cos^2\theta)(Sin^2\theta)} + \frac{Sin^2\theta}{(1 + Sin^2\theta)(Cos^2\theta)}) \times Sin^2\theta Cos^2\theta((1+Cos2θ)(Sin2θ)Cos2θ+(1+Sin2θ)(Cos2θ)Sin2θ)×Sin2θCos2θ

= (\frac{Cos^4\theta + Cos^4\theta Sin^2\theta + Sin^4\theta + Sin^4\theta Cos^2\theta}{(1 + Cos^2\theta)(Sin^2\theta)(1 + Sin^2\theta)(Cos^2\theta)}) \times Sin^2\theta Cos^2\theta((1+Cos2θ)(Sin2θ)(1+Sin2θ)(Cos2θ)Cos4θ+Cos4θSin2θ+Sin4θ+Sin4θCos2θ)×Sin2θCos2θ

= \frac{Cos^4\theta + Cos^2\theta Sin^2\theta (Cos^2\theta + Sin^2\theta) + Sin^4\theta }{1 + Cos^2\theta + Sin^2\theta + Cos^2\theta Sin^2\theta}1+Cos2θ+Sin2θ+Cos2θSin2θCos4θ+Cos2θSin2θ(Cos2θ+Sin2θ)+Sin4θ

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