Question

$\frac{sin \alpha }{1 - cos \alpha } + \frac{tan \alpha }{1 + cos \alpha } = sec \: \alpha \ \times cosec \: \alpha \: + \: cot \: \alpha$
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Answer 1

$\huge\text{\underline{Answer}}$

$\sf{\underline{Step by step explanation}}$

$\bold{\frac{sin \alpha }{1 - cos \alpha } + \frac{tan \alpha }{1 + cos \alpha } = sec \: \alpha \ \times cosec \: \alpha \: + \: cot \: \alpha }$

L. H. S= $\bold{\frac{sin \alpha }{1 - cos \alpha } + \frac{tan \alpha }{1 + cos \alpha } }$

$\implies$ $\bold{\frac{sin \alpha }{1 - cos \alpha } + \frac{ \frac{sin \alpha }{cos \alpha } }{1 + cos \alpha } }$

$\implies$ $\bold{[\frac{sin \alpha }{1 - cos \alpha } + \frac{sin \alpha }{cos \alpha (1 + cos \alpha )} ]}$

$\implies$ $\bold{= sin \alpha [ \frac{1}{1 - cos} + \frac{1}{cos \alpha (1 + cos \alpha )}] }$

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

$\boxed{\sf{1 - {cos }^{2} \alpha = {sin }^{2} \alpha }}$

$\implies$ $\bold{=sin \alpha [\frac{cos \alpha + {cos \alpha }^{2} + 1 - cos \alpha }{cos \alpha {sin \alpha }^{2} } ] }$

$\implies$ $\bold{=\frac{1 + {cos \alpha }^{2} }{cos \alpha \: sin \alpha } }$

$\implies$ $\bold{= \frac{1}{cos \alpha \: sin \alpha \: } + \frac{ {cos \alpha }^{2} }{cos \alpha \: sin \alpha } }$

$\implies$ $\bold{= sec \alpha \: cosec \alpha \: + \frac{cos \alpha }{sin \alpha} }$

$\implies$ $\bold{ sec \alpha \: cosec \alpha \: + cot \alpha }$

Answer 2

Answer:

1−cosα

sinα

+

1+cosα

tanα

=secα ×cosecα+cotα

L. H. S= \bold{\frac{sin \alpha }{1 - cos \alpha } + \frac{tan \alpha }{1 + cos \alpha } }

1−cosα

sinα

+

1+cosα

tanα

\implies⟹ \bold{\frac{sin \alpha }{1 - cos \alpha } + \frac{ \frac{sin \alpha }{cos \alpha } }{1 + cos \alpha } }

1−cosα

sinα

+

1+cosα

cosα

sinα

\implies⟹ \bold{[\frac{sin \alpha }{1 - cos \alpha } + \frac{sin \alpha }{cos \alpha (1 + cos \alpha )} ]}[

1−cosα

sinα

+

cosα(1+cosα)

sinα

]

\implies⟹ \bold{= sin \alpha [ \frac{1}{1 - cos} + \frac{1}{cos \alpha (1 + cos \alpha )}] }=sinα[

1−cos

1

+

cosα(1+cosα)

1

]

\implies⟹ \bold{= sin \alpha [ \frac{cos \alpha (1 + cos \alpha ) + 1 - cos \alpha }{cos \alpha (1 - {cos \alpha }^{2})}]}=sinα[

cosα(1−cosα

2

)

cosα(1+cosα)+1−cosα

]

\boxed{\sf{1 - {cos }^{2} \alpha = {sin }^{2} \alpha }}

1−cos

2

α=sin

2

α

\implies⟹ \bold{=sin \alpha [\frac{cos \alpha + {cos \alpha }^{2} + 1 - cos \alpha }{cos \alpha {sin \alpha }^{2} } ] }=sinα[

cosαsinα

2

cosα+cosα

2

+1−cosα

]

\implies⟹ \bold{=\frac{1 + {cos \alpha }^{2} }{cos \alpha \: sin \alpha } }=

cosαsinα

1+cosα

2

\implies⟹ \bold{= \frac{1}{cos \alpha \: sin \alpha \: } + \frac{ {cos \alpha }^{2} }{cos \alpha \: sin \alpha } }=

cosαsinα

1

+

cosαsinα

cosα

2

\implies⟹ \bold{= sec \alpha \: cosec \alpha \: + \frac{cos \alpha }{sin \alpha} }=secαcosecα+

sinα

cosα

\implies⟹ \bold{ sec \alpha \: cosec \alpha \: + cot \alpha }secαcosecα+cotα