Question

if cosec theta+ cot theta=k then prove that cos theta =k^2-1/K^2+1​

Answer 1

Answer:

See the explanation

Step-by-step explanation:

Given  

           $\cos ec\theta+\cot\theta=k$

          $\Rightarrow \dfrac{1}{\sin\theta}+\dfrac{\cos\theta}{\sin\theta}=k$

           $\Rightarrow \dfrac{1+\cos\theta}{\sin\theta}=k\\\\\Rightarrow 1+\cos\theta=k\sin\theta$

Square both sides

          $\Rightarrow (1+\cos\theta)^2=(k\sin\theta)^2\\\Rightarrow 1+2\cos\theta+\cos^2\theta=k^2\sin^2\theta\\\Rightarrow 1+2\cos\theta+\cos^2\theta=k^2(1-\cos^2\theta)\\\Rightarrow (1+k^2)\cos^2\theta+2\cos\theta+(1-k^2)=0\\\\\Rightarrow \cos\theta=\dfrac{-2\pm \sqrt{(2)^2-4(1+k^2)(1-k^2)}}{2(1+k^2)}$

                       $=\dfrac{-2\pm \sqrt{4-4(1-k^4)}}{2(1+k^2)}=\dfrac{-2\pm \sqrt{4k^4}}{2(1+k^2)}\\\\=\dfrac{-2\pm 2k^2}{2(1+k^2)}=-1,\dfrac{k^2-1}{k^2+1}$

But for $\cos\theta=-1$, given equation is not defined

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