If 1 + s e c α + t a n α 2 t a n α = λ , then 1 + t a n α /2 2 t a n α /2 is equal to

1 + s e c α + t a n α 2 t a n α = 1 + c o s α + s i n α 2 s i n α = ( 1 + t 2 ) + ( 1 − t 2 ) + 2 t 2 ⋅ 1 + t 2 2 t ⋅ ( 1 + t 2 ) = 1 + t 2 t where t = tan 2 α . So the value is λ .

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MathematicsTrigonometric Ratios & IdentitiesMiscellaneous/MixedMedium2 minQB

MathematicsMediumsingle choice

If $\frac{2 t a n α}{1 + s e c α + t a n α} = λ$ , then $\frac{2 t a n α /2}{1 + t a n α /2}$ is equal to

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$\frac{2 t a n α}{1 + s e c α + t a n α} = \frac{2 s i n α}{1 + c o s α + s i n α} = \frac{2 \cdot \frac{2 t}{1 + t ^{2}}}{( 1 + t ^{2} ) + ( 1 - t ^{2} ) + 2 t} \cdot (1 + t^{2}) = \frac{2 t}{1 + t}$ where $t = tan \frac{α}{2}$ . So the value is $λ$ .

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

⏱ 2mℹ️ Source: QB

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