Question

If sin theta = a/b then prove that ( sec thetha +tan thetha ) = √b+a/b-a class 10 ka question h bhagwan apka bala kare
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Answer 1

Given, sinA = a/b

Therefore, cos²A = 1 - sin²A

=> cos²A = 1 - a²/b²

=> cosA = √(b² - a²)/b

In the question: secA + tanA

= 1/cosA + sinA/cosA

= (1 + sinA)/cosA

= (1 + a/b) / (√b²-a²)/b

= [ (b + a)/b ] / [ √(b² - a²)/b ]

= (b + a) / √(b² - a²)

= √(b + a)² / √(b + a)(b - a)

= √(b + a) / √(b - a)

Answer 2

Given :-

If sinθ = a/b

To Prove :-

secθ + tanθ = √b + a/b - a

Solution :-

Here,

LHS = secθ + tanθ

RHS = √b + a/b - a

secθ = 1/cos θ

tanθ = sinθ/cosθ

1/cosθ + sinθ/cosθ

1 + sinθ/cosθ

• cosθ = √(1 - sin²θ)

1 + sinθ/√(1 - sin²θ)

1 + (a/b)/√[1 - (a/b)²]

(b + a/b)/√(b² - a²/b²)

b + a/√(b + a)√(b - a)

√(b + a)/√(b - a)

RHS = √(b + a)/√(b - a)

Hence,Proved