Question

sin theta/sec theta +tan theta -1 +cos theta/cosec theta + cot theta-1=1 .

Answer 1

We know that

$sec \: \theta = \frac{1}{cos \: \theta} \\ \\ tan \: \theta = \frac{sin \: \theta}{cos \: \theta} \\ \\ cosec \: \theta = \frac{1}{sin \: \theta} \\ \\ cot \: \theta = \frac{cos\: \theta}{sin \: \theta}$
put these values in the expression

$=\frac{sin \: \theta}{ \frac{1}{cos \: \theta} + \frac{sin \: \theta}{cos \: \theta} - 1 } + \frac{cos\: \theta}{ \frac{1}{sin \: \theta} + \frac{cos \: \theta}{sin\: \theta} - 1 } \\ \\= \frac{sin \: \theta}{ \frac{1 + sin\theta - cos \: \theta}{cos \: \theta} } + \frac{cos\: \theta}{ \frac{1 + cos \: \theta - sin \: \theta}{sin \: \theta} } \\ \\= \frac{sin \: \theta \: cos \: \theta}{1 + sin \: \theta - cos \: \theta} + \frac{sin \: \theta \: cos \: \theta}{1 + cos \: \theta - sin \: \theta} \\ \\ = \frac{sin \: \theta \: cos \: \theta(1 + cos \: \theta - sin \: \theta) + sin \: \theta \: cos \: \theta \: (1 + sin \: \theta - cos \: \theta)}{ (1 + sin \: \theta - cos \: \theta)(1 + cos\: \theta - sin \: \theta)} \\ \\ = \frac{sin \: \theta \: cos \: \theta(1 + cos \: \theta - sin \: \theta + 1 + sin \: \theta - cos \: \theta)}{1 + cos \: \theta - sin\theta + sin \: \theta + sin \: \theta \: cos \: \theta - {sin}^{2} \theta - cos \: \theta - {cos}^{2}\theta + sin\theta \: cos \: \theta) } \\ \\ = \frac{2sin \: \theta \: cos \: \theta}{1 - ( {sin}^{2}\theta + {cos}^{2}\theta) + 2sin \: \theta \: cos \: \theta } \\ \\ = \frac{2sin \: \theta \: cos \: \theta}{1 - 1+ 2sin \: \theta \: cos \: \theta } \\ \\ = \frac{2sin \: \theta \: cos \: \theta}{ 2sin \: \theta \: cos \: \theta } \\ \\ = 1$
So
$\frac{sin \: \theta}{sec \: \theta + tan \: \theta - 1} + \frac{cos \: \theta}{cosec \: \theta + cot\: \theta - 1} = 1$

Hence proved

Answer 2

Answer:

We have to prove that,

$\frac{\sin \theta}{\sec \theta + \tan \theta-1}+\frac{cos \theta}{\cosec \theta+\cot\theta - 1}=1$

L.H.S.

$=\frac{\sin \theta}{\sec \theta + \tan \theta-1}+\frac{cos \theta}{\cosec \theta+\cot\theta - 1}$

$=\frac{\sin \theta}{\frac{1}{\cos \theta} + \frac{\sin\theta}{\cos \theta}-1}+\frac{cos \theta}{\frac{1}{\sin \theta}+\frac{\cos \theta}{\sin\theta} - 1}$

( ∵ sec x = 1/cos x, cosec x = 1/sin x, tan x = sinx/cos x, cot x = cos x / sin x )

$=\frac{\sin \theta \cos \theta}{1+ \sin\theta-\cos\theta }+\frac{\sin \theta cos \theta}{1+\cos\theta - \sin \theta}}$

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

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

( ∵ (a+b)(a-b) = a² - b² )

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

$=\frac{2\sin \theta \cos \theta}{2\sin \theta \cos \theta}$

( ∵ sin² x + cos² x = 1 )

= 1

= R.H.S

Hence, proved...