Question

If θ is an acute angle such that$tan^{2}\Theta=\frac{8}{7}$, then the value of $\frac{(1+sin\Theta)(1-sin\Theta)}{(1+cos\Theta)(1-cos\Theta)}$ is
(a)$\frac{7}{8}$
(b)$\frac{8}{7}$
(c)$\frac{7}{4}$
(d)$\frac{64}{49}$

Answer 1

SOLUTION :  

The correct option is (a) : 7/8

Given : tan² θ  = 8/7

tan θ = √8/7

In right angle ∆ ,  

tan θ =  perpendicular/base = √8/7

perpendicular = √8 , base = √7

Hypotenuse = √( perpendicular)² + (Base)²

[By Pythagoras theorem]

Hypotenuse = √ (√8)² + (√7)² = √(8 + 7) = √15

Hypotenuse = √15  

sin θ = perpendicular / hypotenuse = √8/√15

cos θ = base/ hypotenuse = √7/√15

The value of : (1 + sin θ)(1 - sin θ) / ( 1 + cos θ) (1 - cos θ )

= [(1 + √8/√15) (1 - √8/√15)] / [(1 + √7/√15) (1 - √7/√15)]

= (1² - (√8/√15)²) /  (1² - (√7/√15)²)

[(a + b ) (a -b) = a² - b²]

= (1 - 8/15)/(1 - 7/15)

= [(15 - 8)/15] / [(15 -7)/15]

= (7/15) / (8/15)

= 7/15 × 15/8

= ⅞

(1 + sin θ)(1 - sin θ) / ( 1 + cos θ) (1 - cos θ ) = ⅞ .  

Hence, the value of  (1 + sin θ)(1 - sin θ) / ( 1 + cos θ) (1 - cos θ ) is ⅞.

HOPE THIS ANSWER WILL HELP YOU…

Answer 2

Option ( A ) is correct.

Explanation :

Here I am using A instead of theta.

Given tan² A = 8/7 ---( 1 )

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We know the algebraic identity:

( x + y )( x - y ) = x² - y²

and

Trigonometric identities:

cos²A + sin²A = 1

i ) 1 - sin² A = cos²A

ii ) 1 - cos²A = sin²A

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Value of

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

= ( 1 - sin²A )/( 1 - cos² A )

= ( cos²A )/( sin²A )

= cot²A

= 1/tan²A

= 1/( 8/7 )

= 7/8

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