Question

prove that tan2a_tan2b = cos2b_cos2a÷cos2b×cos2a= sin2a_since ÷cos2b×cos2a.( 2 is a square is tan, cos, sin. ​

Answer 1

Thus, $\tan^2A-\tan^2 B=\dfrac{\cos^2B-\cos^2A}{\cos^2B\cos^2A} =\dfrac{\sin^2A-\sin^2B}{\cos^2B\cos^2A}$, proved.

Step-by-step explanation:

To prove that, $\tan^2A-\tan^2 B=\dfrac{\cos^2B-\cos^2A}{\cos^2B\cos^2A} =\dfrac{\sin^2A-\sin^2B}{\cos^2B\cos^2A}$.

L.H.S. = $\tan^2A-\tan^2 B$

= $\dfrac{\sin^2A}{\cos^2A} -\dfrac{\sin^2B}{\cos^2B}$

Taking LCM of numerator, we get

= $\dfrac{\sin^2A\cos^2B-\sin^2B\cos^2A}{\cos^2A\cos^2B}}$

Using the trigonometric identity,

$\sin^2A+\cos^2A=1$

= $\dfrac{(1-\cos^2 A)\cos^2B-(1-\cos^2 B)\cos^2A}{\cos^2A\cos^2B}}$

= $\dfrac{\cos^2B-\cos^2 A\cos^2B-\cos^2A+\cos^2 A\cos^2B}{\cos^2A\cos^2B}}$

= $\dfrac{\cos^2B-\cos^2A}{\cos^2A\cos^2B}}$

= M.H.S.

= $\dfrac{1-\sin^2B-(1-\sin^2A)}{\cos^2A\cos^2B}}$

= $\dfrac{1-\sin^2B-1+\sin^2A}{\cos^2A\cos^2B}}$

= $\dfrac{\sin^2A-\sin^2B}{\cos^2A\cos^2B}}$

= R.H.S., proved.

Thus, $\tan^2A-\tan^2 B=\dfrac{\cos^2B-\cos^2A}{\cos^2B\cos^2A} =\dfrac{\sin^2A-\sin^2B}{\cos^2B\cos^2A}$, proved.