Question

Prove that sin theta +1-costheta /cos theta -1+sin theta =1+sin theta /cos theta

Answer 1

Answer:

Step-by-step explanation:

Answer 2

Answer:

Here we want to prove,

$\frac{ \sin( \theta) + 1 - \cos(\theta) }{ \cos(\theta) - 1 + \sin(\theta) } = \frac{1 + \sin(\theta) }{ \cos(\theta) }$

So,

L.H.S,

$\frac{ \sin( \theta) + 1 - \cos(\theta) }{ \cos(\theta) - 1 + \sin(\theta) } \\ = \frac{( \sin( \theta) + 1 - \cos(\theta) )(1 + \sin(\theta) ) \cos(\theta) }{ \cos(\theta) - 1 + \sin(\theta)(1 + \sin(\theta) ) \cos(\theta) }$

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

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

$= \frac{ [\sin( \theta) \cos(\theta) + \cos(\theta) - {cos}^{2}\theta ] (1 + \sin(\theta) }{ [\sin(\theta) \cos(\theta) + \cos(\theta) - {cos}^{2}\theta \ ] cos(\theta) }$

$= \frac{1 + \sin(\theta) }{ \cos(\theta) }$ = R.H.S

It is prove that L.H.S = R.H.S

Here applied formula,

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

Some more important Trigonometry formulas,

$sin(x) = \cos(\frac{\pi}{2} - x) \\ \tan(x) = \cot(\frac{\pi}{2} - x) \\ \sec(x) = \csc(\frac{\pi}{2} - x) \\ \cos(x) = \sin(\frac{\pi}{2} - x) \\ \cot(x) = \tan(\frac{\pi}{2} - x) \\ \csc(x) = \sec(\frac{\pi}{2} - x)$

know more about Trigonometry,

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