Question

trigonometry x. Please provide solution ​

Answer 1

Solution:

Here I am using A instead of theta.

Given sinA = a/b ----(1)

cos²A = 1-sin²A

= 1-(a/b)²

= (b²-a²)/b²

cosA = √(b²-a²)/b ---(2)

Now ,

secA+tanA

= 1/cosA + sinA/cosA

= (1+sinA)/cosA

= (1+a/b)/[(√b²-a²)/b

= (b+a)/(√b²-a²)

= √(b+a)(b+a)/√(b+a)(b-a)

= (√b+a)/(√b-a)

Therefore,

secA+tanA = (√b+a)/(√b-a)

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