If \(2\frac{{{{\cos }^2}x - {{\sec }^2}x}}{{{{\tan }^2}x}} = \) a + b cos 2 x, then a, b = ?

Question

Question

This question was previously asked in

SSC CGL 2022 Tier-I Official Paper (Held On : 09 Dec 2022 Shift 3)

Answer 1

Given:

\(2\frac{{{{\cos }^2}x - {{\sec }^2}x}}{{{{\tan }^2}x}} = a+b\cos2x\)

Formula Used:

\(\cos2x=2\cos^2x-1\) or \(\cos^2x=\frac{\cos2x+1}{2}\)

\(a^2-b^2=(a+b)(a-b)\)

Calculation:

Simplifying L.H.S:

⇒\(2\frac{{{{\cos }^2}x - \frac{1}{\cos^2x}}}{{{\frac{\sin^2x}{\cos^2x}}}} \)

⇒\(2\frac{\frac{\cos^4x-1}{\cos^2x}}{\frac{\sin^x}{\cos^2x}}\)

⇒\(2\frac{(\cos^2x)^2-1^2}{\sin^2x}\)

Applying property \(a^2-b^2=(a+b)(a-b)\)

⇒\(2\frac{(\cos^2x+1)(\cos^2x-1)}{\sin^2x}\)

⇒\(2\frac{(\cos^2x+1)(-\sin^2x)}{\sin^2x}\)

⇒\(-2(\cos^2x+1)\)

Applying formula of \(\cos2x\)

⇒\(-2(\frac{\cos2x+1}{2}+1)\)

⇒\(-2(\frac{\cos2x+3}{2})\)

⇒\(-3-\cos2x\)

Now, comparing the L.H.S with R.H.S we get:

\(a=-3\) and \(b=-1\)

∴ The value of a, b is -3, -1.

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