Question

Sin theta + cos theta =a and sec theta + csc theta =b , then b(a²-1) =​

Answer 1

Answer

Given, secθ+cscθ=b and sinθ+cosθ=a

⇒

sinθ

1

+

cosθ

1

=b

⇒

sinθcosθ

sinθ+cosθ

=b

⇒

sinθcosθ

a

=b

⇒sinθcosθ=

b

a

--- [1]

Now consider, sinθ+cosθ=a

Squaring on both the sides we get,

(sinθ+cosθ)

2

=a

2

⇒sin

2

θ+cos

2

θ+2sinθcosθ=a

2

⇒1+

b

2a

=a

2

----[From1]

⇒

b

(b+2a)

=a

2

⇒(b+2a)=a

2

b

⇒2a=a

2

b−b

⇒2a=b(a

2

−1

Hence, the answer is 2a

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Answer 2

Given

$\sf\ sin\theta+cos\theta=a\\ \\ \sf\ sec\theta+csc\theta= b$

To Find :-

We have to find the value of b(a²-1)

Solution

By putting the given values

$:\implies\sf\ \ b(a^2-1)\\ \\ \\ :\implies\sf\ \ sec\theta+csc\theta\big\{(sin\theta+cos\theta)^2-1)\big\}\\ \\ \\ \bullet\sf\ sec\theta=\dfrac{1}{cos\theta}\ \ ;\ csc\theta=\dfrac{1}{sin\theta}\\ \\ \\ :\implies\sf\ \dfrac{1}{cos\theta}+\dfrac{1}{sin\theta}\big\{sin^2\theta+cos^2\theta+2sin\theta cos\theta-1\big\}\\ \\ \\ \bullet\sf\ \ sin^2\theta+cos^2\theta=1\\ \\ \\ :\implies\sf\dfrac{sin\theta+cos\theta}{sin\theta\ cos\theta}\big\{\cancel{1}+2sin\theta\ cos\theta \cancel{-1}\big\}\\ \\ \\ :\implies\sf\ \dfrac{sin\theta+cos\theta}{\cancel{sin\theta cos\theta}}\times 2\cancel{sin\theta cos\theta}\\ \\ \\ :\implies\sf\ \ 2(sin\theta+cos\theta)$

$\underline{\bigstar{\blue{\sf\ \ b(a^2-1)= 2(sin\theta+cos\theta)}}}$