Question

Please help me with this trigonometry problem

Answer 1

Ok I’m too lazy to write theta so imma write a instead.
1/sec2a+1/cosec2a
=cos2a+sin2a
=1

Please mark me as brainliest.

Answer 2

$\sf{ \frac{1}{ {sec}^{2} \theta } = \frac{1}{sec \: \theta} \times \frac{1}{sec \: \theta} = cos \theta \times cos \theta } \\ \\ = \sf{ {cos}^{2} \theta}$

$\sf{ \frac{1}{ {cosec}^{2} \theta } = \frac{1}{cosec \theta} \times \frac{1}{cosec \theta} = sin \theta \times sin \theta } \\ \\ \sf{ = sin {}^{2} \theta }$

Now,

substitute in the question

$\sf{ {cos}^{2} \theta + {sin}^{2} \theta} \\ = \boxed{ \boxed{ \bf{1}}}$