# Show that (cosecθ-sinθ)(secθ-cosθ)=1/tanθ+cotθ

2
by ruhitheace

2015-03-23T11:13:04+05:30

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LHS
=(1/sin teta -sin teta) (1/cos teta - cos teta)
=(1-sin²teta/sin teta) (1-cos²teta/cos teta)
=(cos²teta/sinteta)(sin²teta/cos teta)
after cancellation remaining term is
=cos teta × sin teta -------------- 1
RHS
=1/tan teta +cot teta
=1/ [sin teta/ cos teta] + [cos teta/sin teta]
=1/ [sin²teta +cos²teta/sin teta×cos teta]
=1/ [1/sin teta × cos teta]
=1 × sin teta ×cos teta/1
=sin teta × cos teta -------------- 2
from 1 and 2 for both LHS and RHS the result came is same
so LHS = RHS
i.e, (cosec teta - sin teta) (sec teta - cos teta)
2015-03-23T11:21:44+05:30

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LHS=(1/sinФ - sinФ) (1/cosФ -cosФ)
=((1-sin²Ф)/sinФ) ((1-cos²Ф)/cosФ)
=(cos²Ф/sinФ) (sin²Ф/cosФ)
=cosФsinФ ----1

RHS = 1/(sinФ/cosФ +cosФ/sinФ)
=1/((sin²Ф+cos²Ф)/(sinФcosФ))
=1/(1/cosФ)(1/sinФ)
=cosФsinФ----2

LHS = RHS
(cosecФ-sinФ)(secФ-cosФ)=1/tanФ+cotФ