Question

show that 1+sinA+cosA/1+cosA-sinA=1+sinA/cosA.​

Answer 1

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Q) Show that

$\sf \: \frac{1 + sin \: a + cos \: a}{1 + cos \: a - sin \: a} = \frac{1 + sin \: a}{cos \: a}$

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Take LHS ,

$\rm \: \frac{1 + sin \: a + cos \: a}{1 + cos \: a - sin \: a}$

$\mapsto \rm \: \frac{(1 + cos \: a) + sin \: a}{(1 + cos \: a) - sin \: a}$

Multiply Neumerator and denominator by the conjugate of denominator

$\mapsto \rm \: \frac{(1 + cos \: a) + sin \: a}{(1 + cos \: a) - sin \: a} \times \frac{(1 + cos \: a) + sin \: a}{(1 + cos \: a) + sin \: a}$

$\mapsto \rm \: \frac{ { \{(1 + cos \: a) + sin \: a \}}^{2} }{ {(1 + cos \: a)}^{2} - {sin}^{2} \: a}$

$\mapsto \rm \: \frac{ {(1 + cos \: a)}^{2} + {sin}^{2} \: a + 2(1 + cos \: a)(sin \: a) }{1 + {cos}^{2} \: a + 2cos \: a - {sin}^{2} \: a }$

$\mapsto \rm \: \frac{1 + \underline{{cos}^{2} \: a } + 2cos \: a+ \underline{ {sin}^{2} \: a }+ 2sin \: a(1 + cos \: a)}{ \cancel{sin}^{2} \: a + {cos}^{2} \: a + {cos}^{2} \: a + 2cos \: a - \cancel {sin}^{2} \: a}$

In the above step ,

We changed , 1 to sin² a + cos² a .

$\mapsto \rm \: \frac{2 + 2cos \: a + 2sin \: a(1 + cos \: a)}{2 {cos}^{2} \: a+ 2cos \: a}$

$\mapsto \rm \: \frac{ \cancel2 \{1 + cos \: a + sin \: a(1 + cos \: a) \}}{ \cancel2( {cos}^{2} \: a + cos \: a)}$

$\mapsto \rm \: \frac{ \cancel{(1 + cos \: a)}(1 + sin \: a)}{cos \: a \cancel{(1 + cos \: a)} }$

$\mapsto \rm \: \frac{1 + sin \: a}{cos \: a}$

As you can see , it is equal to RHS .

Hence Proved !!

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★ Know More :

• cos² x + sin² x = 1
• 1 + tan² x = sec² x
• cot² x + 1 = cosec² x
• $\sf \: sin \: \theta = \frac{perpendicular}{hypotenuse \: }$
• $\sf \: cos \theta = \frac{base}{hypotenuse}$
• $\sf \: tan \theta = \frac{perp}{base}$