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Question Number 105285 by bemath last updated on 27/Jul/20

$$simplify\frac{\sin \theta }{1+\cot \theta }+\frac{\cos \theta }{1+\tan \theta }?$$

Commented by bemath last updated on 27/Jul/20

$$thankyou$$

Commented by som(math1967) last updated on 27/Jul/20

$$\frac{\sin \theta }{1+\frac{\cos \theta }{\sin \theta }}+\frac{\cos \theta }{1+\frac{\sin \theta }{\cos \theta }}$$$$=\frac{{\sin }^2\theta }{\sin \theta +\cos \theta }+\frac{{\cos }^2\theta }{\cos \theta +\sin \theta }$$$$=\frac{1}{\sin \theta +\cos \theta }[\because {\sin }^2\theta +{\cos }^2\theta =1]$$

Answered by bobhans last updated on 27/Jul/20

$$\iff \frac{\sin {}^{2}x}{\sin x+\cos x}+\frac{\cos {}^2}{\cos x+\sin x}=\frac{1}{\sin x+\cos x}$$$$=\frac{1}{\sqrt{2}\cos (x-\frac{\pi }{4})}=\frac{\sec (x-\frac{\pi }{4})}{\sqrt{2}}▹$$

Commented by malwaan last updated on 27/Jul/20

$$\frac{1}{sinx+cosx}=\frac{1}{\sqrt{2}(sinx\times \frac{1}{\sqrt{2}}+cosx\times \frac{1}{\sqrt{2}})}$$$$=\frac{1}{\sqrt{2}(sinxcos\frac{\pi }{4}+cosxsin\frac{\pi }{4})}$$$$=\frac{1}{\sqrt{2}sin(x+\frac{\pi }{4})}=\frac{cosec(x+\frac{\pi }{4})}{\sqrt{2}}$$

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