Question

If tanθ =8/15, then evaluate (1+sinθ)(2-2sinθ)/(2+2cosθ)(1-cosθ)

Answer 1

tanθ = 8/15

(1 + sinθ)(2 - 2sinθ)/(2 + 2cosθ)(1 - cosθ)

= {(1 + sinθ)2(1 - sinθ)}/{2(1 + cosθ)(1 - cosθ)}

= (1 + sinθ)/(1 - sinθ)/(1 + cosθ)(1 - cosθ)

= (1 - sin²θ )/(1 - cos²θ) [ from, (a -b)(a + b) = a² -b²]

= cos²θ/sin²θ [ from sin²θ + cos²θ = 1 ]

= 1/{sin²θ/cos²θ}

= 1/tan²θ

put , tanθ = 8/15

= 1/{8/15}²

= 15²/8²

= 225/64 [ans]

Answer 2

Hi ,

Here I am using ' A ' instead of theta.

tanA = 8/15 ----( 1 )

i ) ( 1 + sinA )( 2 - 2SinA )

= 2( 1 + sinA )( 1 - sinA )

= 2( 1² - sin²A )

= 2cos²A ---( 2 )

ii ) ( 2 + 2 cosA )( 1 - cosA )

= 2( 1 + cosA )( 1 - cosA )

= 2 ( 1² - cos² A )

= 2sin²A ---( 3 )

Now ,

[(1+sinA)(2-2sinA)]/[(2+2cosA)(1-cosA)]

= [2cos²A ]/[ 2sin²A] from ( 2) and ( 3 ),

= cos²A/sin²A

= cot²A

= 1/tan²A

= 1/(8/15)²

= 15²/8²

= 225/64

I hope this helps you.

: )