Question

f b cos theta is equal to 8 then prove that cos theta + cot theta is equal to root b + a upon B minus a​

Answer 1

Answers

Given bcos\theta = abcosθ=a

\implies cos\theta = \frac{a}{b}⟹cosθ=ba ----(1)

Now ,

LHS = cosec\theta+cot\thetaLHS=cosecθ+cotθ

=\frac{1}{sin\theta}+\frac{cos\theta}{sin\theta}sinθ1+sinθcosθ

=\frac{(1+cos\theta)}{sin\theta}sinθ(1+cosθ)

= \frac{\sqrt{(1+cos\theta)^{2}}}{\sqrt{sin^{2}\theta}}sin2θ(1+cosθ)2

=\frac{\sqrt{(1+cos\theta)^{2}}}{\sqrt{1-cos^{2}\theta}}1−cos2θ(1+cosθ)2

=\frac{\sqrt{(1+cos\theta)^{2}}}{\sqrt{(1-cos\theta)}\times \sqrt{1+cos\theta)}}(1−cosθ)×1+cosθ)(1+cosθ)2

=\sqrt{\frac{(1+cos\theta)}{(1-cos\theta)}}(1−cosθ)(1+cosθ)

= \sqrt{\frac{1+\frac{a}{b}}{1-\frac{a}{b}}}1−ba1+ba

/* from (1) */

=\sqrt{\frac{\frac{b+a}{b}}{\frac{b-a}{b}}}bb−abb+a

= \sqrt{\frac{b+a}{b-a}}b−ab+a

Therefore,

cosec\theta+cot\thetacosecθ+cotθ =\sqrt{\frac{b+a}{b-a}}b−ab+a