Question

please solve my mathematics problem​

Answer 1

Step-by-step explanation:

√sec²θ+cosec²θ

=√1+tan²θ+1+cot²θ

=√tan²θ+cot²θ+2

=√(tanθ+cotθ)²

=tanθ+cotθ

Answer 2

Answer:

taking LHS

Using

sec²θ = 1 + tan²θ

cosec²θ = 1 + cot²θ

$\sqrt{ { \sec }^{2} \theta + { \cosec}^{2}\theta }$

$= \sqrt{1 + { \tan}^{2} \theta + 1 + { \cot}^{2} \theta }$

$= \sqrt{ { \tan}^{2} \theta + 2 + { \cot}^{2} \theta}$

2 can write as 2tanθcotθ

Because ,

tanθ×cotθ = 1 because they are reciprocal to each other

$= \sqrt{ { \tan}^{2} \theta + 2 \tan \theta \cot \theta + { \cot}^{2} \theta }$

It is in the form of a² + 2ab +b² = (a + b)²

$= \sqrt{ {( \tan \theta + \cot \theta)}^{2} } \\ = \tan \theta + \cot \theta$

LHS = RHS

Hence proved