Question

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

Answer 1

Answer:

please make me brainlist

Step-by-step explanation:

The given equation is:

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

1+sinθ

cosθ

=

cosθ

1−sinθ

Taking the LHS of the above equation, we have

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

1+sinθ

cosθ

=\frac{cos\theta}{1+sin{\theta}}{\times}\frac{1-sin\theta}{1-sin{\theta}}

1+sinθ

cosθ

×

1−sinθ

1−sinθ

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

1−sin

2

θ

cosθ(1−sinθ)

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

cos

2

θ

cosθ(1−sinθ)

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

cosθ

1−sinθ

=RHS

Answer 2

Trigonometry:-

Trigonometry

Question:-

.

$prove \: that \: \\ \frac{cos \theta}{1 - sin \theta} = \frac{1 + sin \theta}{cos \theta}$

Solution:-

$take \: lhs \\ \\ \frac{cos \theta}{1 - sin \theta} \\ \\ multipy \: and \: divide \: (1 + sin \theta) \\ \\ = \frac{cos \theta(1 + sin \theta}{(1 - sin \theta)(1 + sin \theta)} \\ \\ = \frac{cos \theta(1 + sin \theta)}{1 - {sin}^{2} \theta } \\ \\ = \frac{cos \theta(1 + sin \theta)}{ {cos}^{2} \theta} \\ \\ = \frac{1 + sin \theta}{cos \theta} \\ \\ hence \: proved \:$

More Information:-

Trigon metric Identities

sin²θ + cos²θ = 1

sec²θ - tan²θ = 1

csc²θ - cot²θ = 1

Trigometric relations

sinθ = 1/cscθ

cosθ = 1 /secθ

tanθ = 1/cotθ

tanθ = sinθ/cosθ

cotθ = cosθ/sinθ

Trigonmetric ratios

sinθ = opp/hyp

cosθ = adj/hyp

tanθ = opp/adj

cotθ = adj/opp

cscθ = hyp/opp

secθ = hyp/adj