Question

If sin A - cos A = 0 for some acute angle A, then find the value of

\(2tan^ 2A +\frac {1}{cosec^2 A} + 1\)

Answer 1

sin A = cos A ⇒ A = 45°

\(2tan ^ 2 A +\frac {1}{cosec ^ 2 A}+ 1 = 2tan ^2 A + sin ^2 A + 1\)

\(=2tan ^ 2 45 ° +sin ^ 2 45 ° + 1\)

\(=2(1)^2 +(\frac {1}{\sqrt {2}})^1 +1\)

\(=2 + \frac {1}{2} + 1 =\frac {7}{2}\)