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Question Number 199817 by ajfour last updated on 09/Nov/23

Answered by ajfour last updated on 09/Nov/23

$${\{a^2+\frac{1}{4}(1-\tan {}^2\theta )-\frac{\sin \theta }{2(1+\sin \theta )}\}}^2$$$$=a\{a+\frac{\sin \theta }{2\cos \theta (1+\sin \theta )}\}$$$$b=\frac{\sin \theta }{2\cos \theta (1+\sin \theta )}$$

Answered by ajfour last updated on 09/Nov/23

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