Question

please any one can solve it in nice way.. question no.. 2

Answer 1

Hey !!!

cos¢(1 /1-sin¢ - 1/1+sin¢)

cos¢ {1+sin¢ -(1 -sin¢) / (1²- sin²¢) } [just lcm ]

cos¢ { 1 + sin¢ - 1 + sin¢/cos²¢ }

cos¢ ( 2sin¢/cos²¢ } ✓•°• 1 - sin²¢ = cos²¢

= 2sin¢/cos¢ ✓

= 2tan¢

now given in the form of ktan¢

so , 2tan¢ = ktan¢

=> k = 2 Answer

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Hope it helps you !!!

@Rajukumar111

Answer 2

= cosΘ(1/1-sinΘ  -  1/1+sinΘ)    

= cosΘ(1+sinΘ-(1-sinΘ)/(1-sinΘ)(1+sinΘ)

= cosΘ(1+sinΘ-1+sinΘ/(1-sin²Θ)

= cosΘ(2sinΘ/cos²Θ)                               

(as 1-sin²Θ = cos²Θ)

= 2(sinΘ/2cosΘ)           
 
= 2(tanΘ)

Hence k = 2
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