Logo
Jeeify
MCQEasyJEE 2025Trigonometric Ratios & Identities

JEE Mathematics 2025 Question with Solution

The value of (sin⁡70∘)(cot⁡10∘cot⁡70∘−1)(\sin 70^\circ)(\cot 10^\circ \cot 70^\circ - 1) is:

  • A

    11

  • B

    00

  • C

    3/23/2

  • D

    2/32/3

Answer

Correct answer:A

Step-by-step solution

Standard Method

Given: Evaluate (sin⁡70∘)(cot⁡10∘cot⁡70∘−1)(\sin 70^\circ)(\cot 10^\circ \cot 70^\circ - 1).

Find: The value of the given trigonometric expression.

Using

cot⁡Acot⁡B−1=cos⁡Acos⁡Bsin⁡Asin⁡B−1=cos⁡Acos⁡B−sin⁡Asin⁡Bsin⁡Asin⁡B=cos⁡(A+B)sin⁡Asin⁡B\cot A \cot B - 1 = \frac{\cos A \cos B}{\sin A \sin B} - 1 = \frac{\cos A \cos B - \sin A \sin B}{\sin A \sin B} = \frac{\cos(A+B)}{\sin A \sin B}

put A=10∘A = 10^\circ and B=70∘B = 70^\circ.

Then

cot⁡10∘cot⁡70∘−1=cos⁡80∘sin⁡10∘sin⁡70∘\cot 10^\circ \cot 70^\circ - 1 = \frac{\cos 80^\circ}{\sin 10^\circ \sin 70^\circ}

So the expression becomes

(sin⁡70∘)(cos⁡80∘sin⁡10∘sin⁡70∘)=cos⁡80∘sin⁡10∘(\sin 70^\circ)\left(\frac{\cos 80^\circ}{\sin 10^\circ \sin 70^\circ}\right) = \frac{\cos 80^\circ}{\sin 10^\circ}

Now use the complementary angle identity

cos⁡80∘=sin⁡10∘\cos 80^\circ = \sin 10^\circ

Therefore

cos⁡80∘sin⁡10∘=sin⁡10∘sin⁡10∘=1\frac{\cos 80^\circ}{\sin 10^\circ} = \frac{\sin 10^\circ}{\sin 10^\circ} = 1

Therefore, the value of the expression is 11. The correct option is A.

Identity-Based Expansion

Given: (sin⁡70∘)(cot⁡10∘cot⁡70∘−1)(\sin 70^\circ)(\cot 10^\circ \cot 70^\circ - 1)

Find: Its numerical value.

Write cotangent in terms of sine and cosine:

cot⁡10∘=cos⁡10∘sin⁡10∘,cot⁡70∘=cos⁡70∘sin⁡70∘\cot 10^\circ = \frac{\cos 10^\circ}{\sin 10^\circ}, \qquad \cot 70^\circ = \frac{\cos 70^\circ}{\sin 70^\circ}

Then

(sin⁡70∘)(cot⁡10∘cot⁡70∘−1)=(sin⁡70∘)(cos⁡10∘cos⁡70∘sin⁡10∘sin⁡70∘−1)(\sin 70^\circ)(\cot 10^\circ \cot 70^\circ - 1) = (\sin 70^\circ)\left(\frac{\cos 10^\circ \cos 70^\circ}{\sin 10^\circ \sin 70^\circ} - 1\right) =(sin⁡70∘)(cos⁡10∘cos⁡70∘−sin⁡10∘sin⁡70∘sin⁡10∘sin⁡70∘)= (\sin 70^\circ)\left(\frac{\cos 10^\circ \cos 70^\circ - \sin 10^\circ \sin 70^\circ}{\sin 10^\circ \sin 70^\circ}\right)

Using

cos⁡xcos⁡y−sin⁡xsin⁡y=cos⁡(x+y)\cos x \cos y - \sin x \sin y = \cos(x+y)

we get

=(sin⁡70∘)(cos⁡80∘sin⁡10∘sin⁡70∘)= (\sin 70^\circ)\left(\frac{\cos 80^\circ}{\sin 10^\circ \sin 70^\circ}\right)

Cancel sin⁡70∘\sin 70^\circ:

=cos⁡80∘sin⁡10∘= \frac{\cos 80^\circ}{\sin 10^\circ}

Since

cos⁡80∘=sin⁡10∘\cos 80^\circ = \sin 10^\circ

we obtain

=1= 1

Hence, the correct answer is A.

Common mistakes

  • A common mistake is to assume cot⁡10∘cot⁡70∘=1\cot 10^\circ \cot 70^\circ = 1 because the angles look complementary. This is wrong because the identity is tan⁡θ=cot⁡(90∘−θ)\tan \theta = \cot(90^\circ-\theta), not that the product of two cotangents of complementary angles equals 11. First convert carefully or expand using sine and cosine.

  • Students often forget the identity cos⁡xcos⁡y−sin⁡xsin⁡y=cos⁡(x+y)\cos x \cos y - \sin x \sin y = \cos(x+y) and incorrectly combine terms. This changes the numerator and leads to a wrong result. Keep the sign pattern exact before applying the sum formula.

  • Another mistake is not using the complementary angle identity cos⁡80∘=sin⁡10∘\cos 80^\circ = \sin 10^\circ at the last step. Without this substitution, the simplification remains incomplete. Always check whether the final ratio contains complementary trigonometric functions.

Practice more Trigonometric Ratios & Identities questions

Get unlimited AI-adaptive practice, mastery tracking, and an AI tutor that explains every step — free to start.

Start free trialMore on Trigonometric Ratios & Identities

Related questions