Advertisement

Prove the Following. Tan θ S E C θ + 1 = S E C θ − 1 Tan θ - Geometry

Question

Prove the following.
\[\frac{\tan\theta}{sec\theta + 1} = \frac{sec\theta - 1}{\tan\theta}\]

Solution

\[\frac{\tan\theta}{\sec\theta + 1}\]
\[ = \frac{\tan\theta}{\sec\theta + 1} \times \frac{\sec\theta - 1}{\sec\theta - 1}\]
\[ = \frac{\tan\theta\left( \sec\theta - 1 \right)}{\sec^2 \theta - 1}\]
\[ = \frac{\tan\theta\left( \sec\theta - 1 \right)}{\tan^2 \theta} \left( 1 + \tan^2 \theta = \sec^2 \theta \right)\]
\[ = \frac{\sec\theta - 1}{\tan\theta}\]

  Is there an error in this question or solution?
Advertisement

APPEARS IN

Advertisement
Prove the Following. Tan θ S E C θ + 1 = S E C θ − 1 Tan θ Concept: Application of Trigonometry.
Advertisement
Share
Notifications

View all notifications
Login
Create free account


      Forgot password?
View in app×