Question

prove 1+sin theta divided by cos theta+ cos theta divided by 1+sin theta =2 sec theta?​

Answer 1

Step-by-step explanation:

= $2sec\theta$

To Prove :-

$\dfrac{1 + sin \theta}{cos\theta} + \dfrac{cos\theta}{1 + sin\theta} = 2sec \theta$

Solution :-

$\dfrac{1 + sin \theta}{cos\theta} + \dfrac{cos\theta}{1 + sin\theta}$

Taking L.C.M :-

$\dfrac{(1 + sin \theta)^{2} }{cos\theta(1 + sin \theta)} + \dfrac{ {cos }^{2} \theta }{cos\theta(1 + sin \theta)}$

$\dfrac{(1 + sin \theta) ^{2} + {cos}^{2} \theta }{cos \theta +cos \theta \: sin \theta }$

$\dfrac{1 + {sin}^{2} \theta + 2sin \theta + {cos}^{2} \theta }{cos \theta + cos \theta \: sin \theta}$

$\dfrac{1 + {sin}^{2} \theta + + {cos}^{2} \theta + \: 2sin \theta }{cos \theta + cos \theta \: sin \theta}$

we know that :-

sin^2A + cos^2A = 1

$\dfrac{1 + 1 + 2sin \theta}{cos \theta(1 + sin \theta)}$

$\dfrac{2 + 2sin \theta}{cos \theta(1 + sin \theta)}$

$\dfrac{2( 1+ sin \theta}{cos \theta(1 + sin \theta)}$

$\dfrac{2}{cos \theta}$

$2 \times \dfrac{1}{cos \theta}$

we know that 1/cosA = secA

$2 \times sec \theta$

$= 2sec \theta$

Hence proved that :-

$\dfrac{1 + sin \theta}{cos\theta} + \dfrac{cos\theta}{1 + sin\theta} = 2sec \theta$