Question

prove that above question ​

Answer 1

Answer:

$\frac{cos \alpha }{(1 - tan \alpha) } - \frac{ {sin}^{2} \alpha }{(sin \alpha - cos \alpha )} \\ \\ = \frac{cos \alpha }{(1 - \frac{sin \alpha }{cos \alpha } )} - \frac{ {sin}^{2} \alpha }{(sin \alpha - cos \alpha) } \\ \\ = \frac{cos \alpha }{ \frac{cos \alpha - sin \alpha }{cos \alpha } } - \frac{ {sin}^{2} \alpha }{(sin \alpha - cos \alpha )} \\ \\ = \frac{ {cos}^{2} \alpha }{(cos \alpha - sin \alpha) } - \frac{ {sin}^{2} \alpha }{(sin \alpha - cos \alpha )} \\ \\ = \frac{ {cos}^{2} \alpha }{(cos \alpha - sin \alpha )} + \frac{ {sin}^{2} }{(cos \alpha - sin \alpha )} \\ \\ = \frac{ {cos}^{2} \alpha + {sin}^{2} \alpha }{cos \alpha - sin \alpha } \\ \\ = \frac{1}{cos \alpha - sin \alpha } \\ \\ .....(multiplying \: numerator \: and \: denominator \: by \\ \: (cos \alpha + sin \alpha ) \\ \\ = \frac{1}{(cos \alpha - sin \alpha) } \times \frac{cos \alpha + sin \alpha }{cos \alpha + sin \alpha } \\ \\ = \frac{cos \alpha + sin \alpha }{ {cos}^{2} \alpha - {sin}^{2} \alpha } \\ \\ = cos \alpha + sin \alpha ...... = r.h.s$

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Answer 2

Answer:

(1−tanα)

cosα

−

(sinα−cosα)

sin

2

α

=

(1−

cosα

sinα

)

cosα

−

(sinα−cosα)

sin

2

α

=

cosα

cosα−sinα

cosα

−

(sinα−cosα)

sin

2

α

=

(cosα−sinα)

cos

2

α

−

(sinα−cosα)

sin

2

α

=

(cosα−sinα)

cos

2

α

+

(cosα−sinα)

sin

2

=

cosα−sinα

cos

2

α+sin

2

α

=

cosα−sinα

1

.....(multiplyingnumeratoranddenominatorby

(cosα+sinα)

=

(cosα−sinα)

1

×

cosα+sinα

cosα+sinα

=

cos

2

α−sin

2

α

cosα+sinα

=cosα+sinα......=r.h.s