Question

Given cot teeta =7/8,then evaluate (1+sin q) /cos q​

Answer 1

Step-by-step explanation:

Solution(i):

(1+cosθ)(1−cosθ)

(1+sinθ)(1−sinθ)

We have, a

2

−b

2

=(a+b)(a−b)

Similarly,

(1−sin

2

θ)=(1+sinθ)(1−sinθ)

(1−cos

2

θ)=(1+cosθ)(1−cosθ)

Therefore,

(1+cosθ)(1−cosθ)

(1+sinθ)(1−sinθ)

=

(1−cos

2

θ)

(1−sin

2

θ)

=

(1−(

113

7

)

2

)

(1−(

113

8

)

2

)

=

(113−49)

(113−64)

=

64

49

Solution(ii):

Given,

cotθ=

8

7

cot

2

θ=(

8

7

)

2

=

64

49

Answer 2

Answer:

The Question is typed only Half, the complete question is

Q. Given cot theta = 7/8, then evaluate

(i) (1 + sin q) (1 - sin q)/(1 + cos q) (1 - cos q)

Sol:- Given: cot θ = Adjacent/Oppsoite = AB/BC = 7/8

Let: q = θ

∴ (1 + sin θ) (1 - sin θ)/(1 + cos θ) (1 - cos θ) --- Eq (1)

∴ By Pythagoras theorem,

= (BC)² = (AB)² + (AC)²

= (BC)² = (7)² + (8)²

            = 49 + 64

⇒ (BC)² = 113 ⇒ BC = √113

∴ sin θ = Opposite/Hypotenuse = AC/BC = 8/√113

  cosθ = Adjacent/Hypotenuse = AB/BC = 7/√113

⇒ Putting the values of sinθ and cosθ in Eq (1),

= $\frac{(1 + \frac{8}{\sqrt{113} } ) (1-\frac{8}{\sqrt{113} } )}{(1+\frac{7}{\sqrt{113}})(1-\frac{7}{\sqrt{113}})}$  [∵ (a + b) (a - b) = a² - b²]

Here, a = 1 and b = $\frac{8}{\sqrt{113} }$ (Numerator)

         a = 1 and b = $\frac{7}{\sqrt{113} }$  (Denominator)

= $\frac{1^2 - (\frac{8}{\sqrt{113} } )^2} {{1^2 - (\frac{7}{\sqrt{113} } )^2}} = \frac{1-\frac{8^2}{\sqrt{113^2} }}{1-\frac{7^2}{\sqrt{113^2} }}$

= $\frac{1-\frac{64}{113} }{1-\frac{49}{113} } = \frac{\frac{113-64}{113} }{\frac{113-49}{113} }$ ⇒ $\frac{49}{64}$

∴ Required answer is 49/64