Name: Trigonometric Ratios of Complementary Angles
Uploaded: 2019-07-19
Duration: 9 min 3 s
Description: What is the connection between the trigonometric ratios of an angle and its complementary angle? This video not only gives you that but helps you derive them...

Question

Prove the following identity:

\frac{\sec A + \tan A - 1}{\tan A - \sec A + 1} = \frac{\cos A}{1 - \sin A}

Accepted Answer

We have, LHS $= \frac{\sec A + \tan A - 1}{\tan A - \sec A + 1}$

$= \frac{(\sec A + \tan A) - (\sec^{2} A - \tan^{2} A)}{\tan A - \sec A + 1}$ [Write $\sec^{2} θ - \tan^{2} θ$ for $1$ in the numerator]

$= \frac{(\sec A + \tan A) - (\sec A + \tan A) (\sec A - \tan A)}{\tan A - \sec A + 1} [∵ (a^{2} - b^{2}) = (a + b) (a - b)]$

$= \frac{(\sec A + \tan A) \{1 - (\sec A - \tan A)\}}{\tan A - \sec A + 1}$

$= \frac{(\sec A + \tan A) (\tan A - \sec A + 1)}{(\tan A - \sec A + 1)}$

$= \sec A + \tan A$

$= \frac{1}{\cos A} + \frac{\sin A}{\cos A} [∵ \sec θ = \frac{1}{\cos θ}, \tan θ = \frac{\sin θ}{\cos θ}]$

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

$= \frac{(1 + \sin A) (1 - \sin A)}{\cos A (1 - \sin A)}$ [Multiply both the numerator and the denominator by $(1 - \sin A)$ ]

$= \frac{1 - \sin^{2} A}{\cos A (1 - \sin A)} [∵ (a^{2} - b^{2}) = (a + b) (a - b)]$

$= \frac{\cos^{2} A}{\cos A (1 - \sin A)} [∵ \sin^{2} θ + \cos^{2} θ = 1]$

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

$=$ RHS

Hence, proved.

Prove the following identity:
$\frac{\sec A + \tan A - 1}{\tan A - \sec A + 1} = \frac{\cos A}{1 - \sin A}$

Important Questions on Trigonometric Identities

Learn and Prepare for any exam you want !

Prove the following identity: secA+tanA-1tanA-secA+1=cosA1-sinA

Important Questions on Trigonometric Identities

Learn and Prepare for any exam you want !

Prove the following identity:
$\frac{\sec A + \tan A - 1}{\tan A - \sec A + 1} = \frac{\cos A}{1 - \sin A}$