﻿ If \tan \theta =\sqrt{\dfrac{1-a}{a} } \, (0 < a <1), simplify the expression: \dfrac{\sin^2 \theta }{a+\cos

#### Question

If \tan \theta =\sqrt{\dfrac{1-a}{a} } \, (0 < a <1), simplify the expression:

\dfrac{\sin^2 \theta }{a+\cos \theta }+\dfrac{\sin^2 \theta }{a-\cos \theta }

Collected in the board: Trigonometry

Steven Zheng posted 3 years ago

\dfrac{\sin^2 \theta }{a+\cos \theta }+\dfrac{\sin^2 \theta }{a-\cos \theta }

=\dfrac{a\sin^2 \theta -\sin^2 \theta \cos \theta +a\sin^2 \theta +\sin^2 \theta \cos \theta }{a^2-\cos^2 \theta }

=\dfrac{2a\sin^2 \theta }{a^2-\cos^2 \theta }
(1)

Using Power-Reducing Identities

\sin^2\theta = \dfrac{1-\cos (2\theta)}{2}

\cos^2\theta = \dfrac{1+\cos (2\theta)}{2}

(1) is converted to

\dfrac{a(1-\cos 2\theta)}{a^2-\dfrac{1+\cos 2\theta}{2}} = \dfrac{2a(1-\cos 2\theta)}{2a^2-1-\cos 2\theta }
(2)

Using the double angle identity

\cos 2\alpha = \cos^2\alpha - \sin^2 \alpha = \dfrac{\cos^2\alpha - \sin^2 \alpha }{\cos^2 \alpha+\sin^2\alpha } =\dfrac{1-\tan^2 \alpha }{1+\tan^2\alpha }

\cos 2\alpha = \dfrac{1-\dfrac{1-a}{a} }{1+\dfrac{1-a}{a} } = \dfrac{a-1+a}{a+1-a} =2a-1

Plug in (2) gives

\dfrac{2a(1-\cos 2\theta)}{2a^2-1-\cos 2\theta } = \dfrac{2a(1-2a+1)}{2a^2-1-2a+1} =\dfrac{4(1-a)}{2a-2}=-2

Steven Zheng posted 2 years ago

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