Question

If tan theta + sin theta = m and tan theta – sin theta = n show that m^ 2 – n^ 2 = 4 root mn

Answer 1

HELLO DEAR,

tanθ-sinθ=n

m+n = tanθ+sinθ+tanθ-sinθ=2tanθ

m-n = tanθ+sinθ-tanθ+sinθ=2sinθ

mn = (tanθ+sinθ)(tanθ-sinθ)

   

= tan²θ-sin²θ

m²-n²

=(m+n)(m-n)

=2tanθ.2sinθ

=4sinθtanθ--------(1)

-----------4√mn-----------

=4√(tan²θ-sin²θ)

=4√(sin²θ/cos²θ-sin²θ)

=4√sin²θ(1/cos²θ-1)

=4sinθ√(1-cos²θ)/cos²θ

=4sinθ/cosθ√sin²θ [∵, sin²θ+cos²θ=1]

=4sinθtanθ-----------(1)

from--(1) and----(2)

m²-n² = 4√mn

Hope it helps you dear.......

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Answer 2

Given :

$(\tan\theta +\sin\theta ) = m\\\\(\tan\theta -\sin\theta ) = n$

To prove : $m^2 -n^2 = 4\sqrt{mn}$

Explanation:

$m^2 = \tan^2\theta +\sin^2\theta +2\tan\theta \sin \theta\\\\n^2 = \tan^2\theta +\sin^2\theta - 2\tan\theta \sin \theta\\\\then \\\\m^2 - n^2 = (\tan^2\theta +\sin^2\theta +2\tan\theta \sin \theta)-(\tan^2\theta +\sin^2\theta - 2\tan\theta \sin \theta)\\\\m^2 - n^2 =4 \tan\theta \sin\theta\\\\now\\\\4\sqrt{mn}= 4\sqrt{(\tan\theta +\sin\theta)(\tan\theta -\sin\theta)}\\\\= 4\sqrt{\tan^2\theta-\sin^2\theta}\\\\4\sqrt{\frac{\sin^2\theta}{\cos^2theta}-\sin^2\theta}\\\\=4\sin\theta\sqrt{\sec^2theta-1}$

$\\\\=4\sin\theta\tan \theta$

hence proved

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