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JEE • MATHEMATICS • COMPOUND ANGLES

Compound angles practice · Hard

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Practice format and chapter quiz

These sets use single-correct-answer MCQs. They do not cover every JEE Main or JEE Advanced question format and are not full-length mock papers.

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1 Hard

1 of 5

Application +4 | -1

A > 0,\,\;B > 0

and

A + B = \frac{\pi }{3}

, then the maximum value of

\tan A\,\,\tan B

, is

View answer and explanation

Correct option: A

\frac{1}{3}

Option A (

\frac{1}{3}

) is correct because the maximum value of

\tan A \tan B

is achieved when

A = B = \frac{\pi}{6}

, leading to the value

\frac{1}{3}

2 Hard

2 of 5

Application +4 | -1

\cos \,\,20^\circ - \sin \,\,20^\circ = p

, then

\cos \,\,40^\circ

is equal to

View answer and explanation

Correct option: B

p\,\sqrt {2 - {p^2}}

Option B (

p\sqrt{2 - p^2}

) is correct because by squaring the given equation and manipulating trigonometric identities, we can express

\cos 40^\circ

in terms of

p

3 Hard

3 of 5

Analyzing +4 | -1

\alpha ,\,\;\beta ,\,\;\gamma \in \left( {0,\;\,\pi /2} \right)

, then the value of

\frac{{\sin \,\,\left( {\alpha + \beta + \gamma } \right)}}{{\sin \,\,\alpha + \sin \,\,\beta + \sin \,\,\gamma }}

, is

View answer and explanation

Correct option: A

< 1

Option A (

< 1

) is correct because

\sin \alpha + \sin \beta + \sin \gamma > \sin (\alpha + \beta + \gamma)

when

\alpha, \beta, \gamma \in (0, \pi/2)

4 Hard

4 of 5

Analyzing +4 | -1

\;\alpha ,\,\,\beta ,\,\,\gamma \; \in \left( {0,\frac{\pi }{2}} \right)

, then the value of

\frac{{\sin \left( {\alpha + \beta + \gamma } \right)}}{{\sin \,\,\alpha + \sin \,\,\beta + \sin \,\,\gamma }}

View answer and explanation

Correct option: A

< 1

Option A (

< 1

) is correct because

\sin(\alpha + \beta + \gamma) < \sin \alpha + \sin \beta + \sin \gamma

for

\alpha, \beta, \gamma \in (0, \pi/2)

5 Hard

5 of 5

Analyzing +4 | -1

p

is the product of the sines of angles of a triangle, and

q

the product of their cosines, then tangents of the angles are roots of the equation

View answer and explanation

Correct option: A

q{x^3} - p{x^3} + \left( {1 + q} \right)x - p = 0

Option A (

q{x^3} - p{x^3} + \left( {1 + q} \right)x - p = 0

) is correct because the tangents of the angles are the roots of this equation, derived from Vieta's formulas and trigonometric identities.

Revision notes and key takeaways

Review COMPOUND ANGLES in your textbook or class notes, then use this set to check your understanding.

Key Takeaways

✓ This set contains 5 single-correct-answer MCQs on COMPOUND ANGLES.
✓ Try each question before opening its answer. Read the explanation where one is available.
✓ Sign in to save submitted answers. Revisit incorrect answers alongside your textbook or class notes.

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