Question

if the sin¢=cos¢ find the value of 1+tan^2¢​

Answer 1

guess this is the answer...

Answer 2

$\sin( \alpha ) = \cos( \alpha ) \\ \sin(45) = \frac{1}{ \sqrt{2} } \\ \cos(45) = \frac{1}{ \sqrt{2 } } \\ \sin( \alpha ) = \cos( \alpha ) \\ where \ \: alpha = 45 \\ 1 + \tan {}^{2} ( \alpha ) = \sec {}^{2} ( \alpha ) \\ \\ therefore \: \sec {}^{2} {45} = (\sqrt{2} ) {}^{2} \\ \\ \sec {}^{2} (45) = 2$