7. If cot 0=7/8
- evaluate : (i)
(1 + sin 0) (1 - sin o)
(1 + cos 0) (1 - cos )
(ii) cot? 0
Answers
Answer:
cotθ = 49/64
To find (i):
\Longrightarrow \sf \dfrac{(1+sin\theta)(1-sin\theta)}{(1+cos\theta)(1-cos\theta)}⟹
(1+cosθ)(1−cosθ)
(1+sinθ)(1−sinθ)
⇒ Using (a - b)(a + b) = a² - b² we get,
\Longrightarrow \sf \dfrac{1^2-sin^2\theta}{1^2-cos^2\theta}⟹
1
2
−cos
2
θ
1
2
−sin
2
θ
⇒ Using 1 - sin²θ = cos²θ and 1 - cos²θ = sin²θ we get,
\Longrightarrow \sf \dfrac{cos^2\theta}{sin^2\theta}⟹
sin
2
θ
cos
2
θ
⇒ Using cos²θ/sin²θ = cot²θ we get,
\Longrightarrow \sf cot^2\theta⟹cot
2
θ
⇒ Using cotθ = 7/8 we get,
\Longrightarrow \sf (\dfrac{7}{8})^2⟹(
8
7
)
2
\Longrightarrow \sf \dfrac{49}{64}⟹
64
49
Therefore,
\Longrightarrow \sf \dfrac{(1+sin\theta)(1-sin\theta)}{(1+cos\theta)(1-cos\theta)} = \dfrac{49}{64}⟹
(1+cosθ)(1−cosθ)
(1+sinθ)(1−sinθ)
=
64
49
_____________________
To find (ii): cot²θ
\Longrightarrow \sf (\dfrac{7}{8})^2⟹(
8
7
)
2
\Longrightarrow \sf \dfrac{49}{64}⟹
64
49
hope this will be yours answer up to where I have understand
