Question

Given cot theta =7/8,
then evaluate
1). ( 1+sin theta)(1-sin theta) / (1+cos theta)(1-cos theta)
2). 1+sin theta / cos theta​

Answer 1

★Given:-

• Cotθ = 7/8

★To find:-

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

Formula's we need to know:

✦[a²-b²=(a+b)(a-b)]

✦1-sin²θ = cos²θ

✦1-cos²θ = sin²θ

✦cotθ  = cosθ /sinθ

✦sec²θ = 1+tan²θ

✦tanθ = 1/cotθ

★Solution:-

1.(1+sinθ)(1-sinθ) / (1+cosθ)(1-cosθ)

⇒1²-sin²θ/1²-cos²θ

⇒1-sin²θ/1-cos²θ

⇒ cos²θ/sin²θ

⇒cot²θ

Given that cotθ = 7/8

Therefore,

⇒cot²θ = (7/8)²

         = 49/64

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2.(1+sinθ)/cosθ

⇒1/cosθ + sinθ/cosθ

⇒secθ + tanθ....(1)

Using the formula,

✦sec²θ = 1+tan²θ

⇒secθ = √ 1+(64/49)

           =√113 /7

We now have,

• secθ = √113 /7
• tanθ = 1/cotθ = 8/7

Putting these values in (1),

⇒√113 /7 + 8/7

⇒ √113 +8 /7

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