MHT CET · Maths · Trigonometric Ratios & Identities
\(\frac{\sin A+\sin 7 A+\sin 13 A}{\cos A+\cos 7 A+\cos 13 A}=\)
- A \(\cot 7 A\)
- B \(\tan 6 A\)
- C \(\tan 7 A\)
- D \(\cot 6 A\)
Answer & Solution
Correct Answer
(C) \(\tan 7 A\)
Step-by-step Solution
Detailed explanation
\(\frac{\sin A+\sin 7 A+\sin 13 A}{\cos A+\cos 7 A+\cos 13 A} =\frac{(\sin A+\sin 13 A)+\sin 7 A}{(\cos A+\cos 13 A)+\cos 7 A} \)
\( =\frac{2 \sin 7 A \cos 6 A+\sin 7 A}{2 \cos 7 A \cos 6 A+\cos 7 A} \)
\( =\frac{\sin 7 A(2 \cos 6 A+1)}{\cos 7 A(2 \cos 6 A+1)} \)
\( =\tan 7 A\)
\( =\frac{2 \sin 7 A \cos 6 A+\sin 7 A}{2 \cos 7 A \cos 6 A+\cos 7 A} \)
\( =\frac{\sin 7 A(2 \cos 6 A+1)}{\cos 7 A(2 \cos 6 A+1)} \)
\( =\tan 7 A\)
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