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JEE • MATHEMATICS • TRIGONOMETRY RATIOS

Trigonometry ratios practice · Easy

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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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Showing 211-218 of 218 questions

211 Easy

211 of 218

Application +4 | -1

\cos ^2\left(\frac{\pi}{6}+\theta\right)-\sin ^2\left(\frac{\pi}{6}-\theta\right)=

View answer and explanation

Correct option: A

\frac{1}{2} \cos 2 \theta

Option A (

\frac{{1}}{{2}} \cos 2 \theta

) is correct because using the identity

\cos^2 A - \sin^2 B = \cos(A+B)\cos(A-B)

, the expression simplifies to

\frac{{1}}{{2}} \cos 2 \theta

212 Easy

212 of 218

Application +4 | -1

The value of

\cos 20^{\circ}+\cos 80^{\circ}-\sqrt{3} \cos 50^{\circ}

is equal to -

View answer and explanation

Correct option: A

Option A (0) is correct because by using the sum-to-product trigonometric identity and simplifying, the expression evaluates to 0.

213 Easy

213 of 218

Application +4 | -1

The value of

\sin \left(\frac{-11 \pi}{3}\right)

\frac{\sqrt{3}}{m}

. Value of '

m

' is

View answer and explanation

Correct option: A

Option A (2) is correct because given \sin x = \frac{{-2

\sqrt{{6}}

}}{{5}} and x in the third quadrant, calculating \cot x results in \frac{{1}}{{2

\sqrt{{6}}

}}.

214 Easy

214 of 218

Understanding +4 | -1

\tan \frac{A}{2}

is equal to

View answer and explanation

Correct option: C

\pm \sqrt{\frac{1-\cos A}{1+\cos A}}

Option C (

\pm \sqrt{\frac{{1-\cos A}}{{1+\cos A}}}}

) is correct because it's a direct application of the half-angle formula for tangent.

215 Easy

215 of 218

Application +4 | -1

\tan ^2 36^{\circ}+k\left(\sin 18^{\circ}+\cos 36^{\circ}\right)=5

, then the value of

k

View answer and explanation

Correct option: C

The correct answer is C (4) because substituting the known values of

\sin 18^{\circ}

and

\cos 36^{\circ}

into the equation and solving for

k

yields

k=4

216 Easy

216 of 218

Application +4 | -1

\sin \,\,\theta + \mathrm{cosec}\;\,\,\theta = 2

then

{\sin ^2}\,\,\theta + \mathrm{cose}{\mathrm{c}^2}\;\,\,\theta

is equal to

View answer and explanation

Correct option: C

\mathrm{2}

Option C is correct because squaring the given equation and simplifying leads to

{\sin ^2}\,\theta + \mathrm{cose}{\mathrm{c}^2}\,\theta = 2

217 Easy

217 of 218

Application +4 | -1

\mathrm{sec}\,\mathrm{\theta \; + }\,\mathrm{tan}\,\mathrm{\theta } = 1

, then root of the equation

\left( {a - 2b + c} \right){x^2}\,\, + \,\,\left( {b - 2c + a} \right)x\,\, + \,\,\left( {c - 2a + b} \right) = 0

View answer and explanation

Correct option: A

\sec \,\,\mathrm{\theta }

Option A (

\sec \theta

) is correct because one root of the quadratic equation is 1, and given the condition

\sec \theta + \tan \theta = 1

, it follows that

\sec \theta = 1

218 Easy

218 of 218

Understanding +4 | -1

\alpha, \beta, \gamma

are the angles which a half ray makes with the positive directions of the axes, then

\sin ^2 \alpha+\sin ^2 \beta+\sin ^2 \gamma

is equal to

View answer and explanation

Correct option: B

Option B (2) is correct because

\sin^2 \alpha + \sin^2 \beta + \sin^2 \gamma = 2

when

\alpha, \beta, \gamma

are the angles a half ray makes with the positive directions of the axes.

Revision notes and key takeaways

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

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