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The value of the expression  1−sin⁡2t1+sin⁡2t×cos⁡t+sin⁡tcos⁡t−sin⁡t\frac{1 - \sin 2t}{1 + \sin 2t} \times \frac{\cos t + \sin t}{\cos t - \
Question

The value of the expression  1−sin⁡2t1+sin⁡2t×cos⁡t+sin⁡tcos⁡t−sin⁡t\frac{1 - \sin 2t}{1 + \sin 2t} \times \frac{\cos t + \sin t}{\cos t - \sin t}​ is:

A.

1 - 2tan(t) / 1 + 2tan(t)

B.

(1 - tan(t)) / (1 + tan(t))

C.

(1 + 2tan(t)) / (1 - 2tan(t))

D.

(1 + tan(t)) / (1 - tan(t))

Correct option is B

Given:  
​      1−sin⁡2t1+sin⁡2t×cos⁡t+sin⁡tcos⁡t−sin⁡t\frac{1 - \sin 2t}{1 + \sin 2t} \times \frac{\cos t + \sin t}{\cos t - \sin t}​​
Formula Used: 
​sin⁡2t=2sin⁡tcos⁡t\sin 2t = 2\sin t \cos t   
sin⁡2t+cos⁡2t=1\sin^2 t + \cos^2t = 1   
Solution:  
​1−sin⁡2t1+sin⁡2t×cos⁡t+sin⁡tcos⁡t−sin⁡t=>sin⁡2t+cos⁡2t−2sin⁡tcos⁡tsin⁡2t+cos⁡2t+2sin⁡tcos⁡t×(cos⁡t+sin⁡t)(cos⁡t+sin⁡t)(cos⁡t−sin⁡t)(cos⁡t+sin⁡t)=>(cos⁡t−sin⁡t)2(cos⁡t+sin⁡t)2×(cos⁡t+sin⁡t)2cos⁡2t−sin⁡2t=>(cos⁡t−sin⁡t)2(cos⁡t−sin⁡t)(cos⁡t+sin⁡t)=>cos⁡t−sin⁡tcos⁡t+sin⁡t Diving it bycos⁡t gives=>1−tan⁡t1+tan⁡t\frac{1 - \sin 2t}{1 + \sin 2t} \times \frac{\cos t + \sin t}{\cos t - \sin t} \\\Rarr \frac{\sin^2 t + \cos^2t - 2\sin t \cos t}{\sin^2 t + \cos^2t + 2\sin t \cos t} \times \frac{(\cos t + \sin t)(\cos t + \sin t)}{(\cos t - \sin t)(\cos t + \sin t)} \\ \Rarr\frac{ (\cos t - \sin t)^2}{\cancel{(\cos t + \sin t)^2}} \times \frac{\cancel{(\cos t + \sin t )^2}}{ \cos^2 t - \sin^2 t} \\ \Rarr \frac{ (\cos t - \sin t)^{ \cancel 2}}{\cancel {(\cos t - \sin t)}(\cos t + \sin t)} \\ \Rarr \frac{ \cos t - \sin t }{ \cos t + \sin t}\\ \text{ Diving it by} \cos t \text{ gives} \\ \Rarr \frac{ 1 - \tan t}{1+\tan t}   
​​​

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