Question

please solve it questions 24........

Answer 1

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Answer 2

Question:

$\dfrac{2tan\theta}{1+{tan}^{2}\theta} = 2sin\theta.cos\theta$

Solution:

LHS = $\dfrac{2tan\theta}{1+{tan}^{2}\theta}$

RHS = $2sin\theta.cos\theta$

LHS = $\dfrac{2 tan\theta}{1+{tan}^{2}\theta}$

$\dfrac{\dfrac{2sin\theta}{cos\theta}}{1+\dfrac{{sin}^{2}\theta}{{cos}^{2}\theta}}$

$=\dfrac{\dfrac{2sin\theta}{cos\theta}}{\dfrac{{cos}^{2}\theta+{sin}^{2}\theta}{{cos}^{2}\theta}}$

$= \dfrac{\dfrac{2sin\theta}{cos\theta}}{\dfrac{1}{{cos}^{2}\theta}}$

$= \dfrac{2sin\theta}{cos\theta} \times \dfrac{{cos}^{2}\theta}{1}$

$= 2sin\theta.cos\theta$ = RHS

Hence, proved.

☆Identities used☆

• $tan\theta = \dfrac{sin\theta}{cos\theta}$

• ${sin}^{2}\theta + cos^{2}\theta = 1$