Question

costheta / 1- sin theta + 1-sintheta /costheta =2sectheta​

Answer 1

Step-by-step explanation:

Consider the L.H.S

\dfrac{\sin \theta -\cos \theta +1}{\sin \theta +\cos \theta -1}

sinθ+cosθ−1

sinθ−cosθ+1

=\left( \dfrac{\sin \theta -\cos \theta +1}{\sin \theta +\cos \theta -1} \right)\times \left( \dfrac{\sin \theta +\cos \theta +1}{\sin \theta +\cos \theta +1} \right)=(

sinθ+cosθ−1

sinθ−cosθ+1

)×(

sinθ+cosθ+1

sinθ+cosθ+1

)

=\left( \dfrac{\sin \theta +1-\cos \theta }{\sin \theta +\cos \theta -1} \right)\times \left( \dfrac{\sin \theta +1+\cos \theta }{\sin \theta +\cos \theta +1} \right)=(

sinθ+cosθ−1

sinθ+1−cosθ

)×(

sinθ+cosθ+1

sinθ+1+cosθ

)

=\dfrac{{{\left( \sin \theta +1 \right)}^{2}}-{{\cos }^{2}}\theta }{{{\left( \sin \theta +\cos \theta \right)}^{2}}-{{1}^{2}}}=

(sinθ+cosθ )

2

−1

2

(sinθ+1)

2

−cos

2

θ

=\dfrac{{{\sin }^{2}}\theta +1+2\sin \theta -{{\cos }^{2}}\theta }{{{\sin }^{2}}\theta +{{\cos }^{2}}\theta +2\sin \theta \cos \theta -1}=

sin

2

θ+cos

2

θ+2sinθcosθ−1

sin

2

θ+1+2sinθ−cos

2

θ

Since, {{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1sin

2

θ+cos

2

θ=1

Therefore,

=\dfrac{1-{{\cos }^{2}}\theta +1+2\sin \theta -{{\cos }^{2}}\theta }{1+2\sin \theta \cos \theta -1}=

1+2sinθcosθ−1

1−cos

2

θ+1+2sinθ−cos

2

θ

=\dfrac{2-2{{\cos }^{2}}\theta +2\sin \theta }{2\sin \theta \cos \theta }=

2sinθcosθ

2−2cos

2

θ+2sinθ

=\dfrac{1-{{\cos }^{2}}\theta +\sin \theta }{\sin \theta \cos \theta }=

sinθcosθ

1−cos

2

θ+sinθ

=\dfrac{{{\sin }^{2}}\theta +\sin \theta }{\sin \theta \cos \theta }=

sinθcosθ

sin

2

θ+sinθ

=\dfrac{\sin \theta +1}{\cos \theta }=

cosθ

sinθ+1

=\dfrac{1}{\cos \theta }+\dfrac{\sin \theta }{\cos \theta }=

cosθ

1

+

cosθ

sinθ

=\sec \theta +\tan \theta =secθ+tanθ

=\left( \sec \theta +\tan \theta \right)\times \left( \dfrac{\sec \theta -\tan \theta }{\sec \theta -\tan \theta } \right)=(secθ+tanθ )×(

secθ−tanθ

secθ−tanθ

)

=\dfrac{{{\sec }^{2}}\theta -{{\tan }^{2}}\theta }{\sec \theta -\tan \theta }=

secθ−tanθ

sec

2

θ−tan

2

θ

We know that

{{\sec }^{2}}\theta -{{\tan }^{2}}\theta =1sec

2

θ−tan

2

θ=1

Therefore,

=\dfrac{1}{\sec \theta -\tan \theta }=

secθ−tanθ

Answer 2

$\frac{cos \theta}{1 - sin \theta} + \frac{1 - sin \theta}{cos \theta} \\ \\ = \frac{ {cos}^{2} \theta + ( {1 - sin \theta})^{2} }{cos \theta(1 - sin \theta)} \\ \\ = \frac{ {cos}^{2} \theta + 1 + {sin}^{2} \theta - 2sin \theta }{cos \theta(1 - sin \theta)} \\ \\ = \frac{2 - 2sin \theta}{cos \theta(1 - sin \theta)} \\ \\ = \frac{2}{cos \theta} \\ \\ = 2sec \theta$