Question

Prove that tan (A+B) x tan (A-B)= cos²B- cos²A by cos² B-sin²A​

Answer 1

Question:-

Prove that $\displaystyle\sf {\tan (A+B)\tan (A-B)=\dfrac {\cos^2B-\cos^2A}{\cos^2B-\sin^2A}}$

Solution:-

$\displaystyle\longrightarrow\sf {LHS}$

$\displaystyle\longrightarrow\sf {\tan (A+B)\tan (A-B)}$

$\displaystyle\longrightarrow\sf {\dfrac {\sin (A+B)\sin (A-B)}{\cos (A+B)\cos (A-B)}}$

$\displaystyle\longrightarrow\sf {\dfrac {[\sin A\cos B+\cos A\sin B][\sin A\cos B-\cos A\sin B]}{[\cos A\cos B-\sin A\sin B][\cos A\cos B+\sin A\sin B]}}$

$\displaystyle\longrightarrow\sf {\dfrac {\sin^2 A\cos^2B-\cos^2A\sin^2 B}{\cos^2A\cos^2 B-\sin^2A\sin^2 B}}$

$\displaystyle\longrightarrow\sf {\dfrac {(1-\cos^2 A)\cos^2B-\cos^2A(1-\cos^2 B)}{(1-\sin^2A)\cos^2 B-\sin^2A(1-\cos^2 B)}}$

$\displaystyle\longrightarrow\sf {\dfrac {\cos^2B-\cos^2 A\cos^2B-\cos^2A+\cos^2A\cos^2 B}{\cos^2B-\sin^2A\cos^2 B-\sin^2A+\sin^2A\cos^2 B}}$

$\displaystyle\longrightarrow\sf {\dfrac {\cos^2B-\cos^2A}{\cos^2B-\sin^2A}}$

$\displaystyle\longrightarrow\sf {RHS}$

Hence Proved!