Show that \(\frac{1}{1+cos(90° -θ)}+\frac1{1-cos(90° – θ)}\) =

1.	Show that \(\frac{1}{1+cos(90° -θ)}+\frac1{1-cos(90° – θ)}\) = 2 cosec2 (90° – θ)
Answer» L.H.S. = \(\frac1{1+sin\,θ}+ \frac1{1-sin\,\theta}\) = \(\frac{1-sin\,\theta+1+sin\,\theta}{(1+sin\,\theta)(1-sin\,\theta)}\) = \(\frac2{1-sin^2\,θ}= \frac2{cos^2\,\theta}\) = 2sec² θ R.H.S.= 2 cosec² (90° – θ) = 2 sec² θ. ∴ L.H.S. = R.H.S

Show that \(\frac{1}{1+cos(90° -θ)}+\frac1{1-cos(90° – θ)}\) = 2 cosec2 (90° – θ)

Answer»

L.H.S. = \(\frac1{1+sin\,θ}+ \frac1{1-sin\,\theta}\) = \(\frac{1-sin\,\theta+1+sin\,\theta}{(1+sin\,\theta)(1-sin\,\theta)}\)

= \(\frac2{1-sin^2\,θ}= \frac2{cos^2\,\theta}\) = 2sec² θ

R.H.S.= 2 cosec² (90° – θ) = 2 sec² θ.

∴ L.H.S. = R.H.S