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Mathematics · Trigonometry Ratios and Identities

JEE Main 2024 — 1 February, Shift 1 — Question 4

If tan⁡A=1x(x2+x+1),tan⁡B=xx2+x+1\tan \mathrm{A}=\frac{1}{\sqrt{x\left(x^{2}+x+1\right)}}, \tan B=\frac{\sqrt{x}}{\sqrt{x^{2}+x+1}} and tan⁡C=(x−3+x−2+x−1)12,\tan C=\left(x^{-3}+x^{-2}+x^{-1}\right)^{\frac{1}{2}}, 0<A,B,C<π2, then A+B<A,B,C<\frac{\pi }{2},\,then \,A+B is equal to

  1. Option A:

    C

    Correct
  2. Option B:

    π−C\pi-C

  3. Option C:

    2π−C2 \pi-C

  4. Option D:

    π2−C\frac{\pi}{2}-C

Answer: A

Step-by-step solution

Finding tan⁡(A+B)\tan (A+B) we get \Rightarrow \tan (A+B)=$$\frac{\tan A+\tan B}{1-\tan A \tan B}=\frac{\frac{1}{\sqrt{x\left(x^{2}+x+1\right)}}+\frac{\sqrt{x}}{\sqrt{x^{2}+x+1}}}{1-\frac{1}{x^{2}+x+1}}

⇒tan⁡(A+B)=(1+x)(x2+x+1)(x2+x)(x)\Rightarrow \tan (\mathrm{A}+\mathrm{B})=\frac{(1+x)\left(\sqrt{x^{2}+x+1}\right)}{\left(x^{2}+x\right)(\sqrt{x})}

tan⁡(A+B)=x2+x+1xx=tan⁡C\tan (A+B)=\frac{\sqrt{x^{2}+x+1}}{x \sqrt{x}}=\tan C

A+B=CA+B=C

Answer key and solution verified before publishing.

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Exam
JEE Main 2024
Paper
1 February, Shift 1
Subject
Mathematics
Chapter
Trigonometry Ratios and Identities
Topic
Trigonometric Ratios of Compound Angles
If tan A =frac 1 sqrt x (x 2 +x+1 ) , tan B=frac √(x) sqrt x 2 +x+1… | JEE Main 2024 PYQ with Solution · DhiX AI