Question

Cos theta upon 1 minus tan
theta + sin square theta upon sin theta minus cos theta simplify it.

Answer 1

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Answer 2

Therefore, The answer is $(sin\theta+cos\theta)$

Step-by-step explanation:

Given,

Simplify, $\frac{cos\theta}{1-tan\theta} +\frac{sin^{2}\theta }{sin\theta-cos\theta}$

∴$\frac{cos\theta}{1-tan\theta} +\frac{sin^{2}\theta }{sin\theta-cos\theta}$

$=\frac{cos\theta}{1-\frac{sin\theta}{cos\theta} } +\frac{sin^{2}\theta }{sin\theta-cos\theta}$

$=\frac{cos^{2} \theta}{cos\theta-sin\theta } -\frac{sin^{2}\theta }{cos\theta-sin\theta}$

$=\frac{cos^{2} \theta-sin^{2} \theta}{cos\theta-sin\theta }$

$=\frac{(cos\theta-sin\theta)(cos\theta+sin\theta)}{cos\theta-sin\theta }$

$=cos\theta+sin\theta$

$=sin\theta+cos\theta$

∴ The answer is $(sin\theta+cos\theta)$