Question

ques.................

Answer 1

◆ Trigonometric Resolutions ◆

Hey !!

Check the attachment.
Hope it helps you :)

Answer 2

Heya User,

--> 7 sin² θ + 3 cos² θ  = 4
=> 3 ( sin² θ + cos² θ ) + 4 sin²θ = 4
=> 3 ( 1 ) + 4 sin²θ = 4   [ sin² θ + cos² θ = 1 ]
=> 4 sin² θ = 4 - 3 = 1

=> sin² θ = 1 / 4
=> sin θ = ± 1 / 2
=> cosec θ = ± 2 

However, cos² θ = [ 1 - sin² θ ] = [ 1 - 1/4 ] = 3/4
            => cos θ = ± √3 / 2
            => sec θ = ± 2√3 / 3 

=> sec θ + cosec θ = ± 2√3 / 3 ± 2  = ± 2 [ 1 + 1/√3 ]

For the sake of scoring better marks --> Plz don't forget to rationalize the denominator --> 
 
--> sec θ + cosec θ = ± $\frac{2 \sqrt{3} [ ( \sqrt{3} + 1) ]}{3}$

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One can utilize the fact that sin 30 = 1/2 or sin ( 210 ) = -1/2

To get --> cosec θ + sec θ = ± $\frac{2 \sqrt{3} [ ( \sqrt{3} + 1) ]}{3}$

^_^ And hence, we are done...