Question

Cot theta ×cot 2theta+cot 2theta×cot 3theta+2=cot theta(cot theta-cot 3theta)

Answer 1

Answer:

The value of \frac{\cot \theta}{\cot \theta-\cot 3 \theta}+\frac{\tan \theta}{\tan \theta-\tan 3 \theta} is 1

Given:

\frac{\cot \theta}{\cot \theta-\cot 3 \theta}+\frac{\tan \theta}{\tan \theta-\tan 3 \theta}

To find:

The value of \frac{\cot \theta}{\cot \theta-\cot 3 \theta}+\frac{\tan \theta}{\tan \theta-\tan 3 \theta}

Solution:

To simplify, \frac{\cot \theta}{\cot \theta-\cot 3 \theta}+\frac{\tan \theta}{\tan \theta-\tan 3 \theta}

Take LCM, the equation becomes,

=\frac{[\cot \theta \times(\tan \theta-\tan 3 \theta)][\tan \theta \times(\cot \theta-\cot 3 \theta]}{(\cot \theta-\cot 3 \theta) \times(\tan \theta-\tan 3 \theta)}

=\frac{[\cot \theta \times(\tan \theta-\tan 3 \theta)][\tan \theta \times(\cot \theta-\cot 3 \theta]}{(\cot \theta \times \tan \theta)-(\cot 3 \theta \times \tan \theta)+(\cot \theta \times \tan \theta)-(\tan 3 \theta \times \cot \theta)}

Since \cot \theta \times \tan \theta=1

=\frac{1-(\cot \theta \times \tan 3 \theta)+1-(\tan \theta \times \cot 3 \theta)}{(1)-(\cot 3 \theta \times \tan \theta)+(1)-(\tan 3 \theta \times \cot \theta)}

=\frac{2-(\cot \theta \times \tan 3 \theta)-(\tan \theta \times \cot 3 \theta)}{(2)-(\cot 3 \theta \times \tan \theta)-(\tan 3 \theta \times \cot \theta)}

Since numerator and denominator are equal, \frac{\cot \theta}{\cot \theta-\cot 3 \theta}+\frac{\tan \theta}{\tan \theta-\tan 3 \theta}=1

Answer 2

Answer:

Cos^2theta/cot^2theta-cos^2theta=3 or, sin^2theta-cos^2theta=3....(1) (because cos^2theta/cot^2theta =sin^2 theta). In the rule of trigonometry sin^2theta+cos^2theta=1 ...(2) we get 1 and 2 .sin^2theta-cos^2theta+sin^2theta+cos^2theta=3+1 or, 2sin^2theta =4 so sin^2theta=2 and cos^2theta=1 ( 1-2) and cot^2 theta= cos^2theta/sin^2theta=1/2

thanks for the answer mate