If tan ( α − β ) = 3 − c o s 2 β s i n 2 β , then

3 − c o s 2 β s i n 2 β = 4 s i n 2 β + 2 c o s 2 β 2 s i n β c o s β = 1 + 2 t a n 2 β t a n β = 1 + 2 t a n 2 β 2 t a n β − t a n β and tan ( α − β ) = 1 + t a n α t a n β t a n α − t a n β . Comparing, tan α = 2 tan β .

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Mathematics - Trigonometric Ratios & Identities Question with Solution | TestHub

MathematicsTrigonometric Ratios & IdentitiesMiscellaneous/MixedMedium2 minQB

MathematicsMediumsingle choice

If $tan (α - β) = \frac{s i n 2 β}{3 - c o s 2 β}$ , then

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Answer:

Solution:

$\frac{s i n 2 β}{3 - c o s 2 β} = \frac{2 s i n β c o s β}{4 s i n ^{2} β + 2 c o s ^{2} β} = \frac{t a n β}{1 + 2 t a n ^{2} β} = \frac{2 t a n β - t a n β}{1 + 2 t a n ^{2} β}$ and $tan (α - β) = \frac{t a n α - t a n β}{1 + t a n α t a n β}$ . Comparing, $tan α = 2 tan β$ .

Stream:JEESubject:MathematicsTopic:Trigonometric Ratios & IdentitiesSubtopic:Miscellaneous/Mixed

⏱ 2mℹ️ Source: QB

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