Given tan A \(\frac { 4 }{ 3 } \) find the other trigonometric ratios of the angle A.

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.

Prove that sec A (1 - sin A) (sec A + tan A) = 1.

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

\(\frac { sin\quad \theta }{ 1+cos\theta } +\frac { 1+cos\theta }{ sin\quad \theta } =2cosec\theta \)

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

If tan A=\(\frac{3}{4}\), then sin A cos A=\(\frac{12}{15}\)

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

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

\(1+\frac { { cot }^{ 2 }\alpha }{ 1+cosex\alpha } =cosec\alpha\)

tan θ+tan(90^o-θ)=secθ sec(90^o-θ)

Find the angle of elevation of the sun when the shadow of a pole h metres high is \(\sqrt3\) h metres long.

If \(\sqrt3\) tan θ=1, then find the value of sin²θ-cos²θ

A ladder 15 metres long just reaches the top of a vertical wall. If the ladder makes an angle of 60^o with the wall, finf the height of the wall.

Simplify (1+tan²θ)(1-sinθ)(1+sinθ)