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MCQMediumJEE 2026Inverse Trigonometric Functions

JEE Mathematics 2026 Question with Solution

Considering the principal values of inverse trigonometric functions, the value of tan⁡ ⁣(2sin⁡−1 ⁣213−2cos⁡−1 ⁣310)\tan\!\left(2\sin^{-1}\!\frac{2}{\sqrt{13}}-2\cos^{-1}\!\frac{3}{\sqrt{10}}\right) is equal to:

  • A

    3356\dfrac{33}{56}

  • B

    −3356-\dfrac{33}{56}

  • C

    1663\dfrac{16}{63}

  • D

    −1663-\dfrac{16}{63}

Answer

Correct answer:A

Step-by-step solution

Standard Method

Given: The expression is

tan⁡(2sin⁡−1(213)−2cos⁡−1(310))\tan\left(2\sin^{-1}\left(\frac{2}{\sqrt{13}}\right)-2\cos^{-1}\left(\frac{3}{\sqrt{10}}\right)\right)

Find: The correct option.

Let

A=sin⁡−1(213)A=\sin^{-1}\left(\frac{2}{\sqrt{13}}\right)

Then

sin⁡A=213,cos⁡A=313,tan⁡A=23\sin A=\frac{2}{\sqrt{13}},\quad \cos A=\frac{3}{\sqrt{13}},\quad \tan A=\frac{2}{3}

Using

tan⁡2A=2tan⁡A1−tan⁡2A\tan 2A=\frac{2\tan A}{1-\tan^2 A}

we get

tan⁡2A=2⋅231−(23)2=431−49=4359=125\tan 2A=\frac{2\cdot\frac{2}{3}}{1-\left(\frac{2}{3}\right)^2}=\frac{\frac{4}{3}}{1-\frac{4}{9}}=\frac{\frac{4}{3}}{\frac{5}{9}}=\frac{12}{5}

Continue with the second angle and combine

Let

B=cos⁡−1(310)B=\cos^{-1}\left(\frac{3}{\sqrt{10}}\right)

Then

cos⁡B=310,sin⁡B=110,tan⁡B=13\cos B=\frac{3}{\sqrt{10}},\quad \sin B=\frac{1}{\sqrt{10}},\quad \tan B=\frac{1}{3}

So

tan⁡2B=2tan⁡B1−tan⁡2B=231−19=2389=34\tan 2B=\frac{2\tan B}{1-\tan^2 B}=\frac{\frac{2}{3}}{1-\frac{1}{9}}=\frac{\frac{2}{3}}{\frac{8}{9}}=\frac{3}{4}

Now use

tan⁡(2A−2B)=tan⁡2A−tan⁡2B1+tan⁡2Atan⁡2B\tan(2A-2B)=\frac{\tan 2A-\tan 2B}{1+\tan 2A\tan 2B}

Therefore

tan⁡(2A−2B)=125−341+125⋅34=48−152020+3620=3356\tan(2A-2B)=\frac{\frac{12}{5}-\frac{3}{4}}{1+\frac{12}{5}\cdot\frac{3}{4}}=\frac{\frac{48-15}{20}}{\frac{20+36}{20}}=\frac{33}{56}

Therefore, the value is 3356\frac{33}{56} and the correct option is A.

Work directly with tangent ratios

Given: The two inverse trigonometric terms.

Find: The value of the tangent expression.

A quick route is to convert each inverse trigonometric quantity into a right-triangle ratio and immediately write tan⁡A\tan A and tan⁡B\tan B. From

sin⁡A=213\sin A=\frac{2}{\sqrt{13}}

we get

tan⁡A=23\tan A=\frac{2}{3}

and from

cos⁡B=310\cos B=\frac{3}{\sqrt{10}}

we get

tan⁡B=13\tan B=\frac{1}{3}

Then compute

tan⁡2A=125,tan⁡2B=34\tan 2A=\frac{12}{5},\quad \tan 2B=\frac{3}{4}

and substitute into the tangent difference formula:

tan⁡(2A−2B)=125−341+125⋅34=3356\tan(2A-2B)=\frac{\frac{12}{5}-\frac{3}{4}}{1+\frac{12}{5}\cdot\frac{3}{4}}=\frac{33}{56}

This works because tangent of a difference depends only on the tangent values of the individual angles. Hence the correct option is A.

Common mistakes

  • A common mistake is taking sin⁡B=110\sin B=\frac{1}{\sqrt{10}} from cos⁡B=310\cos B=\frac{3}{\sqrt{10}}. This is wrong because

    sin⁡2B=1−cos⁡2B=1−910=110\sin^2 B=1-\cos^2 B=1-\frac{9}{10}=\frac{1}{10}

    so sin⁡B=110\sin B=\frac{1}{\sqrt{10}} only after using the principal value range of cos⁡−1\cos^{-1}. Always use the principal-value quadrant before assigning the sign.

  • Students often apply

    tan⁡(α−β)=tan⁡α−tan⁡β1−tan⁡αtan⁡β\tan(\alpha-\beta)=\frac{\tan\alpha-\tan\beta}{1-\tan\alpha\tan\beta}

    which is incorrect. The correct identity is

    tan⁡(α−β)=tan⁡α−tan⁡β1+tan⁡αtan⁡β\tan(\alpha-\beta)=\frac{\tan\alpha-\tan\beta}{1+\tan\alpha\tan\beta}

    Use the plus sign in the denominator for tangent of a difference.

  • Another mistake is substituting directly into the double-angle formula without first finding tan⁡A\tan A and tan⁡B\tan B correctly. For example, using sin⁡A\sin A in place of tan⁡A\tan A leads to the wrong value of tan⁡2A\tan 2A. First convert the inverse trigonometric data into the required basic ratio.

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