Question

Please solve it
Question number 50

Answer 1

Explanation:

We know that:

${ cot (90°-\theta) = tan \theta} \\</p><p>cosec (90°-\theta) = sec \theta\\</p><p>tan (90°-\theta) = cot \theta$

Now

$\frac{cot (90°-\theta)}{tan \theta}+\frac{cosec(90°-\theta)\: sin \theta}{tan (90°-\theta)}= sec^2 \theta\\</p><p>L. H. S. =\frac{cot (90°-\theta)}{tan \theta}+\frac{cosec(90°-\theta)\: sin \theta}{tan (90°-\theta)}$

(Using above properties we find:)

$=\frac{tan \theta}{tan \theta}+\frac{sec\theta\: sin \theta}{cot\theta}\\</p><p>=1+\frac{sec\theta\: sin \theta}{cot\theta}\\</p><p>=1+\frac{tan \theta}{cot\theta}\\</p><p>=1+tan \theta \times tan \theta\\</p><p>=1+tan^2 \theta\\</p><p>=sec^2 \theta\\</p><p>= R. H. S.$

Thus Proved

Answer 2

Answer:

Explanation:

The same doubt i too have .