`(Sec Theta -1 )/( Sec Theta +1) = ( Sin ^2 Theta)/( (1+ Cos Theta )^2)` - Mathematics

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

LHS = $\frac{\sec θ - 1}{\sec θ + 1}$

= $\frac{\frac{1}{\cos θ} - 1}{\frac{1}{\cos θ} + 1}$

= $\frac{\frac{1 - \cos θ}{\cos θ}}{\frac{1 + \cos θ}{\cos θ}}$

= $\frac{1 - \cos θ}{1 + \cos θ}$

= $\frac{(1 - \cos θ) (1 + \cos θ)}{(1 + \cos θ) (1 + \cos θ)} {Dividing the numerator anddenominator by (1 + \cos θ)}$

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

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

= RHS

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Chapter 8: Trigonometric Identities - Exercises 1

Chapter 8 Trigonometric Identities
Exercises 1 | Q 20.1

Prove the following identities, where the angles involved are acute angles for which the expressions are defined:

$\frac{\tan θ}{1 - \cot θ} + \frac{\cot θ}{1 - \tan θ} = 1 + \sec θ \cos e c θ$

[Hint: Write the expression in terms of sinθ and cosθ]

Prove the following identities, where the angles involved are acute angles for which the expressions are defined:

$\frac{\cos A - \sin A + 1}{\cos A + \sin A - 1} = \cos e c A + \cot A$ using the identity cosec² A = 1 cot²A.

if $\cos θ = \frac{5}{13}$ where $θ$ is an acute angle. Find the value of $\sin θ$

Prove the following trigonometric identities.

(cosec θ − sec θ) (cot θ − tan θ) = (cosec θ + sec θ) ( sec θ cosec θ − 2)

Prove the following identities:

$\sqrt{\frac{1 - \cos A}{1 + \cos A}} = \frac{\sin A}{1 + \cos A}$

$\sin^{6} θ + \cos^{6} θ = 1 - 3 \sin^{2} θ \cos^{2} θ$

Write the value of $(1 - \sin^{2} θ) \sec^{2} θ .$

Write the value of $(1 + \tan^{2} θ) (1 + \sin θ) (1 - \sin θ)$

If cosec θ − cot θ = α, write the value of cosec θ + cot α.