Consider the L.H.S.
=√1−sinθ1+sinθ+√1+sinθ1−sinθ
=√1−sinθ√1+sinθ+√1+sinθ√1−sinθ
=1−sinθ+1+sinθ√1+sinθ√1−sinθ
=2√1−sin2θ
=2√cos2θ
=2cosθ
Since, for π2<θ<π
Therefore,
=2−cosθ
=−2cosθ
Hence, proved.
Prove the following trigonometric identities:
Prove that:
(i) √sec θ−1sec θ+1+√sec θ+1sec θ−1=2 cosec θ
(ii) √1+sin θ1−sin θ+√1−sin θ1+sin θ=2 cosec θ
(iii) √1+cos θ1−cos θ+√1−cos θ1+cos θ=2 cosec θ
(iv) sec θ−1sec θ+1=(sin θ1+cos θ)2