Question

Prove the following
secθ+tanθ=cos/1−sinθ[/tex]

Answer 1

Here's the answer

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

$\sf {L. H. S = sec \theta + tan \theta}$

$\sf{= \frac{1}{cos \theta} + \frac{sin \theta}{cos \theta}}$

Because

$\sf{{sec \theta = \frac{ 1}{ cos \theta} \: and \: tan \theta = \frac{sin \theta} { cos \theta}}}$

Therefore, Multiplying (1 - sin theta) with the numerator and denominator ,

Next step will be :

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

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

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

$\sf{= \frac{cos \theta}{1 - sin \theta}}$

= R. H. S

Therefore,

L. H. S = R. H. S

Hence , Proved !

_______________________________

Thanks!!

Answer 2

Hey mate ^_^

=======
Answer:
=======

To prove:

$secθ + tanθ$ = $\frac{cos}{1 - sin θ}$

=====
$Proof:$
=====

$L. H. S. = secθ + tanθ$

= $\frac{1}{cos θ} + \frac{sinθ}{cosθ}$

= $\frac{cosθ}{(cosθ)^{2} } + \frac{cosθ sinθ}{(cosθ)^{2} }$

$= \frac{cosθ + cosθsinθ}{cos^{2}θ}$

Now,

By the identity,

$sin ^{2} θ + cos ^{2} θ = 1$

We get,

$= \frac{cosθ +cosθsinθ}{1 - sin^{2}θ}$

$= \frac{cosθ(1 + sinθ)}{(1 - sinθ)(1 + sinθ)}$

By multiplying and dividing it by $(1+sinθ)$

We get,

$= \frac{cosθ}{1 - sinθ}$

$= R. H. S.$

Hence proved.

#Be Brainly❤️