Question

prove that cos theta /1-sin theta + 1-sin theta/ cos theta = 2 sec theta​

Answer 1

LHS = cos theta / 1-sin theta + 1-sin theta / cos theta

= cos theta * cos theta + 1-sin theta * 1-sin theta / cos theta * 1-sin theta

= cos sq theta + 1+ sin sq theta -2sin theta / cos theta * 1-sin theta

= 2 -2sin theta / cos theta * 1-sin theta

=2(1-sin theta )/ cos theta * 1-sin theta

=2/cos theta

= 2 * 1/cos theta

=2 sec theta

= RHS

Hence proved

Answer 2

Answer:

$\frac{cos \ \theta}{1-sin \theta} +\frac{1-sin\theta}{cos\theta} = 2 sec \theta$ is proved

Step-by-step explanation:

To prove,

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

Solution:

Recall the concepts:

(a-b)² = a²-2ab+b²

sin²θ + cos²θ = 1

$sec \theta = \frac{1}{cos\theta}$

$\frac{cos \ \theta}{1-sin \theta} +\frac{1-sin\theta}{cos\theta}$

Taking LCM as (1-sinθ)cosθ we get

$\frac{cos \ \theta}{1-sin \theta} +\frac{1-sin\theta}{cos\theta}$ =  $\frac{cos^2 \theta + (1-sin\ \theta)^2}{(1-sin \theta)cos\theta}$

Applying the identity (a-b)² = a²-2ab+b² we get

$\frac{cos^2 \theta + (1-sin\ \theta)^2}{(1-sin \theta)cos\theta}$= $\frac{cos^2 \theta + 1+sin^2\ \theta-2sin\theta}{(1-sin \theta)cos\theta}$

applying the identity sin²θ + cos²θ = 1 we get

= $\frac{1+ 1-2sin\theta}{(1-sin \theta)cos\theta}$

= $\frac{2-2sin\theta}{(1-sin \theta)cos\theta}$

= $\frac{2(1-sin\theta)}{(1-sin \theta)cos\theta}$

= $\frac{2}{cos \theta}$

= 2 sec θ( ∵ $sec \theta = \frac{1}{cos\theta}$)

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

Hence proved

#SPJ3