Question

prove that sin theta - cos theta +1/sn theta +cos theta - 1 = 1/sec theta - tan theta​

Answer 1

nswer

Consider the L.H.S

sinθ+cosθ−1

sinθ−cosθ+1

=( lsinθ+cosθ−1sinθ−cosθ+1)×( sinθ+cosθ+1sinθ+cosθ+1)

=(sinθ+cosθ−1sinθ+1−cosθ)×(sinθ+cosθ+1sinθ+1+cos)

= (sinθ+cosθ) 2−1

2(sinθ+1) 2−cos2θ

= sin2θ+cos2θ+2sinθcosθ−1sin 2θ+1+2sinθ−cos 2θ

Since, sin2θ+cos2θ=1

Therefore,= 1+2sinθcosθ−1

1−cos 2θ+1+2sinθ−cos2θ

= 2sinθcosθ2−2cos 2θ+2sinθ

= sinθcosθ1−cos 2θ+sinθ

= sinθcosθsin 2θ+sinθ

=cosθsinθ+1

= cosθ+cosθsinθ

=secθ+tanθ

=(secθ+tanθ)×(secθ−tanθsecθ−tanθ)

= secθ−tanθsec 2θ−tan 2θ

We know that

sec 2θ−tan2θ=1

Therefore,

= secθ−tanθ1

Hence, proved