Question

tan2α−tanα(1+sec2α)=​

Answer 1

ANSWER:

To Find:

• Value of tan2α−tanα(1+sec2α)

Solution:

$:\longrightarrow\tan2\alpha-\tan\alpha(1+\sec2\alpha)\\\\\text{We know that, \tan\theta=\dfrac{\sin\theta}{\cos\theta}\:\:\&\:\:\sec\theta=\dfrac{1}{\cos\theta} . So,}\\\\:\implies\dfrac{\sin2\alpha}{\cos2\alpha}-\dfrac{\sin\alpha}{\cos\alpha}\left(1+\dfrac{1}{\cos2\alpha}\right)\\\\\text{Also, \sin2\theta=2\sin\theta\cos\theta }\\\\:\implies\dfrac{2\sin\alpha\cos\alpha}{\cos2\alpha}-\dfrac{\sin\alpha}{\cos\alpha}\left(\dfrac{\cos2\alpha+1}{\cos2\alpha}\right)\\\\\text{And, \cos2\theta+1=2\cos^2\theta }$

$\text{So,}\\\\:\implies\dfrac{2\sin\alpha\cos\alpha}{\cos2\alpha}-\dfrac{\sin\alpha}{\cos\alpha}\left(\dfrac{2\cos^2\alpha}{\cos2\alpha}\right)\\\\:\implies\dfrac{2\sin\alpha\cos\alpha}{\cos2\alpha}-\dfrac{\sin\alpha\times2\cos^2\alpha}{\cos\alpha\times\cos2\alpha}\\\\:\implies\dfrac{2\sin\alpha\cos\alpha}{\cos2\alpha}-\dfrac{2\sin\alpha\cos^2\alpha}{\cos\alpha\cos2\alpha}$

$:\implies\dfrac{2\sin\alpha\cos^2\alpha-2\sin\alpha\cos^2\alpha}{\cos\alpha\cos2\alpha}\\\\:\implies\dfrac{0}{\cos\alpha\cos2\alpha}\\\\\bf{:\implies 0}$

Formulae Used:

• tanA = (sinA)/(cosA)
• secA = (1)/(cosA)
• sin2A = 2sinAcosA
• cos2A = 2cos²A-1 ⇒ cos2A + 1 = 2cos²A

Answer 2

refer the attachment

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