Question

Prove that:sin theta/sin(90-theta)+ cos theta/ cos (90-theta)=sec theta cosec theta

Answer 1

first to solve this question ,
keep in mind that,
sin (90 - theta)= cos theta
cos(90 - theta)= sin theta
so,
sin theta/sin(90-theta)+cos theta/ cos(90-theta)
= sin theta/ cos theta   +    cos theta/ sin theta
= (sin square theta + cos square theta)/ sin theta. cos theta
= 1/sin theta. cos theta
= cosec theta. sec theta

Answer 2

Answer:

To prove : $\frac{sin \theta}{sin(90-\theta)}+\frac{cos \theta}{cos(90-\theta)}= sec \theta cosec \theta$

Solution :

LHS =  $\frac{sin \theta}{sin(90-\theta)}+\frac{cos \theta}{cos(90-\theta)}$

Identity : $sin (90-\theta ) = cos \theta , cos(90-\theta)=sin \theta$

$\frac{sin \theta}{cos\theta}+\frac{cos \theta}{sin\theta}$

$\frac{sin^2 \theta+cos^2\theta}{cos\theta \times sin\theta}$

We know that $Sin^2\theta +cos^2\theta = 1$

$\frac{1}{cos\theta \times sin\theta}$

Identity : $\frac{1}{cosA}=sec A , \frac{1}{sin A}=cosec A$

So, $sec\theta \times cosec \theta$

Hence $\frac{sin \theta}{sin(90-\theta)}+\frac{cos \theta}{cos(90-\theta)}= sec \theta cosec \theta$