Question

if a nad B are positive actue angles and cos2a=ncos2B-1/n-cos2B(n>1),show that,√n-1tana=√n+1tanB​

Answer 1

Answer:

Step-by-step explanation:

We have,

$\tt{cos(2A)=\dfrac{n\,cos(2B)-1}{n-cos(2B)}}$

$\sf{\implies\,\dfrac{1-tan^{2}(A)}{1+tan^{2}(A)}=\dfrac{n\,\left\{\dfrac{1-tan^{2}(B)}{1+tan^{2}(B)}\right\}-1}{n-\left\{\dfrac{1-tan^{2}(B)}{1+tan^{2}(B)}\right\}}}$

$\sf{\implies\,\dfrac{1-tan^{2}(A)}{1+tan^{2}(A)}=\dfrac{n\,\left(1-tan^{2}(B)\right)-\big(1+tan^{2}(B)\big)}{n\big(1+tan^{2}(B)\big)-\left(1-tan^{2}(B)\right)}}$

$\sf{\implies\,\dfrac{1-tan^{2}(A)}{1+tan^{2}(A)}=\dfrac{n-n\,tan^{2}(B)-1-tan^{2}(B)}{n+n\,tan^{2}(B)-1+tan^{2}(B)}}$

$\sf{\implies\,\dfrac{1-tan^{2}(A)}{1+tan^{2}(A)}=\dfrac{(n-1)-(n+1)\,tan^{2}(B)}{(n-1)+(n+1)\,tan^{2}(B)}}$

Using Componendo and Dividendo

$\sf{\implies\,\dfrac{1-tan^{2}(A)+1+tan^{2}(A)}{1-tan^{2}(A)-1-tan^{2}(A)}=\dfrac{(n-1)-(n+1)\,tan^{2}(B)+(n-1)+(n+1)\,tan^{2}(B)}{(n-1)-(n+1)\,tan^{2}(B)-(n-1)-(n+1)\,tan^{2}(B)}}$$\sf{\implies\,\dfrac{2}{-2tan^{2}(A)}=\dfrac{2(n-1)}{-2(n+1)\,tan^{2}(B)}}$

$\sf{\implies\,\dfrac{1}{tan^{2}(A)}=\dfrac{(n-1)}{(n+1)\,tan^{2}(B)}}$

$\sf{\implies\,(n+1)\,tan^{2}(B)=(n-1)tan^{2}(A)}$

$\sf{\implies\,\sqrt{(n+1)}\,tan(B)=\sqrt{(n-1)}\,tan(A)}$