If θ ={m²-n²}{m²+n²}, then what is θ ?

Correct Answer is : 2mnm2−n2{2mn}{m²-n²}m2−n22mn

Correct Answer is : 2mnm2−n2{2mn}{m²-n²}m2−n22mn

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SSC CPO Tier 1 Exam 10 Dec 2025 Shift 1

Section: Quantitative Aptitude

Question : 141 of 200

Marks: +1, -0

\cos \theta =\frac{m^{2}-n^{2}}{m^{2}+n^{2}}

\tan \theta

Solution:

Concept:

Use the Pythagorean identity

\sin^2 \theta + \cos^2 \theta = 1

to find

\sin \theta

, then

\tan \theta = \frac{\sin \theta}{\cos \theta}

Explanation:

Given:

\cos \theta = \frac{m^{2}-n^{2}}{m^{2}+n^{2}}

Square both sides:

\cos^2 \theta = \frac{(m^{2}-n^{2})^{2}}{(m^{2}+n^{2})^{2}}

Using

\sin^2 \theta = 1 - \cos^2 \theta

\sin^2 \theta = \frac{(m^{2}+n^{2})^{2} - (m^{2}-n^{2})^{2}}{(m^{2}+n^{2})^{2}}

Simplify numerator:

(m^{4}+2m^{2}n^{2}+n^{4}) - (m^{4}-2m^{2}n^{2}+n^{4}) = 4m^{2}n^{2}

Thus

\sin^2 \theta = \frac{4m^{2}n^{2}}{(m^{2}+n^{2})^{2}}

, so

\sin \theta = \frac{2mn}{m^{2}+n^{2}}

(taking positive value).

Now

\tan \theta = \frac{\sin \theta}{\cos \theta} = \frac{2mn/(m^{2}+n^{2})}{(m^{2}-n^{2})/(m^{2}+n^{2})} = \frac{2mn}{m^{2}-n^{2}}

Answer:

\boxed{\frac{2mn}{m^{2}-n^{2}}}

(Option A)

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