Get the solution to "How to simplify [2+tan2xsec2x]−1?"

Conor Hogan

Answered question

2022-12-28

How to simplify

[\frac{2 + \tan^{2} x}{\sec^{2} x}] - 1

mosellasv2

Beginner2022-12-29Added 7 answers

Since,

\sec^{2} x = 1 + \tan^{2} x

, we possess

[\frac{2 + \tan^{2} x}{\sec^{2} x}] - 1

= [\frac{(2 + \tan^{2} x) - \sec^{2} x}{\sec^{2} x}]

= [\frac{2 - (\sec^{2} x - \tan^{2} x)}{\sec^{2} x}]

= \frac{2 - 1}{\sec^{2} x}

= \frac{1}{\sec^{2} x}

= \cos^{2} x

yegumbi4q0

Beginner2022-12-30Added 3 answers

u \sin g t h e t r i g o n o m e t r i c i d e n t i t y

• x 1 + \tan^{2} x = \sec^{2} x

\Rightarrow [\frac{1 + 1 + \tan^{2} x}{\sec^{2} x}] - \frac{\sec^{2} x}{\sec^{2} x}

= \frac{1 + \sec^{2} x}{\sec^{2} x} - \frac{\sec^{2} x}{\sec^{2} x}

= \frac{1}{\sec^{2} x} = \cos^{2} x

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