BITSAT 2021
Exam BITSAT
Year 2021
Subject Mathematics
Chapter Trigonometric Ratios & Identities BITSAT 2021 · Mathematics · Trigonometric Ratios & Identities
If A+B+C=(π)/(2) then
If
A + B + C = π 2 \mathrm{A}+\mathrm{B}+\mathrm{C}=\frac{\pi}{2} A + B + C = 2 π then
Options A tan A tan B + tan B tan C + tan C tan A = 1 \tan A \tan B+\tan B \tan C+\tan C \tan A=1 tan A tan B + tan B tan C + tan C tan A = 1 B cot A + cot B + cot C = cot A cot B cot C \cot A+\cot B+\cot C=\cot A \cot B \cot C cot A + cot B + cot C = cot A cot B cot C C cos 2 A + cos 2 B + cos 2 C = 1 + 4 sin A sin B sin C \cos 2 \mathrm{ A}+\cos 2 \mathrm{ B}+\cos 2 \mathrm{C}=1+4 \sin \mathrm{A} \sin \mathrm{B} \sin C cos 2 A + cos 2 B + cos 2 C = 1 + 4 sin A sin B sin C D All three are correct Correct answer D. All three are correct
Step-by-step solution
B + C = π 2 − A ⇒ tan ( B + C ) = cot A ⇒ tan B + tan C 1 − tan B tan C = cot A ⇒ tan A tan B + tan B tan C + tan C tan A = 1 ⇒ tan A tan B tan C ( cot C + cot A + cot B ) = 1 ⇒ cot A + cot B + cot C = cot A cot B cot C \mathrm{B}+\mathrm{C}=\frac{\pi}{2}-\mathrm{A} \Rightarrow \tan (\mathrm{B}+\mathrm{C})=\cot \mathrm{A}\Rightarrow \frac{\tan \mathrm{B}+\tan \mathrm{C}}{1-\tan \mathrm{B} \tan \mathrm{C}}=\cot \mathrm{A}\Rightarrow \tan \mathrm{A} \tan \mathrm{B}+\tan \mathrm{B} \tan \mathrm{C}+\tan \mathrm{C} \tan \mathrm{A}=1\Rightarrow \tan \mathrm{A} \tan \mathrm{B} \tan \mathrm{C}(\cot \mathrm{C}+\cot \mathrm{A}+\cot \mathrm{B})=1\Rightarrow \cot \mathrm{A}+\cot \mathrm{B}+\cot \mathrm{C}=\cot \mathrm{A} \cot \mathrm{B} \cot \mathrm{C} B + C = 2 π − A ⇒ tan ( B + C ) = cot A ⇒ 1 − t a n B t a n C t a n B + t a n C = cot A ⇒ tan A tan B + tan B tan C + tan C tan A = 1 ⇒ tan A tan B tan C ( cot C + cot A + cot B ) = 1 ⇒ cot A + cot B + cot C = cot A cot B cot C Again
cos 2 A + cos 2 B + cos 2 C = 2 cos ( A + B ) cos ( A − B ) + cos 2 C = 2 cos ( π 2 − C ) cos ( A − B ) + 1 − 2 sin 2 C = 2 sin C [ cos ( A − B ) − sin C ] + 1 = 2 sin C [ cos ( A − B ) − sin π 2 − ( A + B ) ] + 1 \cos 2 \mathrm{ A}+\cos 2 \mathrm{ B}+\cos 2 \mathrm{C}=2 \cos (\mathrm{A}+\mathrm{B}) \cos (\mathrm{A}-\mathrm{B})+\cos 2 \mathrm{C}=2 \cos \left(\frac{\pi}{2}-C\right) \cos (A-B)+1-2 \sin ^{2} C=2 \sin C[\cos (A-B)-\sin C]+1=2 \sin \mathrm{C}\left[\cos (\mathrm{A}-\mathrm{B})-\sin \frac{\pi}{2}-(\mathrm{A}+\mathrm{B})\right]+1 cos 2 A + cos 2 B + cos 2 C = 2 cos ( A + B ) cos ( A − B ) + cos 2 C = 2 cos ( 2 π − C ) cos ( A − B ) + 1 − 2 sin 2 C = 2 sin C [ cos ( A − B ) − sin C ] + 1 = 2 sin C [ cos ( A − B ) − sin 2 π − ( A + B ) ] + 1 = 2 sin C [ cos ( A − B ) − cos ( A + B ) ] + 1 = 4 sin A sin B sin C + 1 \begin{array}{l} =2 \sin C[\cos (A-B)-\cos (A+B)]+1 \\ =4 \sin A \sin B \sin C+1 \end{array} = 2 sin C [ cos ( A − B ) − cos ( A + B )] + 1 = 4 sin A sin B sin C + 1 $$
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