Question

If $cosec\theta +\cot \theta =k$ then $\cos \theta =\frac{k^2-1}{k^2+1}$. is ?

• A. True
• B. False

Answer 1

The correct option is A True

Given,

$cosec\theta +cot\theta =k$

$\frac{1}{sin\theta }+\frac{cos\theta }{sin\theta }=k$

$\frac{1+cos\theta }{sin\theta }=k$

$1+cos\theta =ksin\theta$

Squaring both side,

$(1+cos\theta {)}^2=k^{2}si{n}^2\theta$

$(1+cos\theta {)}^2=k^2(1-co{s}^2\theta )$

$(1+cos\theta ).(1+cos\theta )=k^2(1+cos\theta )(1-cos\theta )$

$1+cos\theta =k^2(1-cos\theta )$

$1+cos\theta =k^2-k^{2}cos\theta$

$1+(1+k^2)cos\theta =k^2$

$(1+k^2)cos\theta =k^2-1$

$cos\theta =\frac{k^2-1}{k^2+1}$

The correct option is A.