Prove that \(\frac { sin\theta -cos\theta +1 }{ sin\theta +cos\theta -1 } =\frac { 1 }{ sec\theta -tan\theta } \) using the identity sec²θ= 1+ tan²θ.

In a right triangle ABC, right-angled at B, if tan A = 1, then verify that 2 sin A cos A = 1.

\(\frac { tan\quad A }{ 1\quad sec\quad A } ​​​​\frac { tan\quad A }{ 1\quad sec\quad A } ​​​​2cosec\quad A\)

(sin ∝+cos ∝)(tan ∝+cot ∝)=sec ∝+cosec ∝

\(\sqrt3\)1 (3-cot30^o)=tan³ 60^o-2sin60^o