What is {{{ x - x}}{{ x + x}}} equal to?

Correct Answer is : 1sec⁡x+tan⁡x{1}{{ x + x}}secx+tanx1

Correct Answer is : 1sec⁡x+tan⁡x{1}{{ x + x}}secx+tanx1

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UPSC CDS 1 2020 Math Paper

Question : 62 of 100

Marks: +1, -0

What is

\sqrt {\frac{{\sec x - \tan x}}{{\sec x + \tan x}}}

equal to?

Solution:

Concept:

Use the identity

\sec^2 x - \tan^2 x = 1

and simplify the expression under the square root.

Explanation:

Start with the given expression:

\sqrt{\frac{\sec x - \tan x}{\sec x + \tan x}}

Multiply numerator and denominator inside the square root by

(\sec x + \tan x)

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

Using the identity

\sec^2 x - \tan^2 x = 1

, we get:

\sqrt{\frac{1}{(\sec x + \tan x)^2}} = \frac{1}{|\sec x + \tan x|}

For typical angles in trigonometry,

\sec x + \tan x > 0

, so the absolute value can be dropped:

\frac{1}{\sec x + \tan x}

Answer:

Option C:

\frac{1}{\sec x + \tan x}

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