if cot B=12/5,prove that tan^2B-sin^2B=sin^4B sec^4B

Open in App

Solution

Suggest Corrections

Similar questions

\frac{12}{5}

{tan}^{2} B - {sin}^{2} B = {sin}^{4} B {sec}^{2} B

t a n^{2} a - t a n^{2} b = \frac{s i n^{2} a - s i n^{2} b}{c o s^{2} a . c o s^{2} b}

t a n^{2} A - t a n^{2} B = \frac{c o s^{2} B - c o s^{2} A}{c o s^{2} B c o s^{2} A} = \frac{s i n^{2} A - s i n^{2} B}{c o s^{2} A c o s^{2} B}

cos B = \frac{12}{5}

{tan}^{2} B - {sin}^{2} B = {sin}^{4} B {sec}^{2} B

\frac{(t a n^{2} A - t a n^{2} B = (s i n^{2} A - s i n^{2} B)}{(C o s^{2} A \times C o s^{2} B)}

Join BYJU'S Learning Program