Respuesta :

SalmonWorldTopper

Given that

Sin θ = a/b

LHS = Sec θ + Tan θ

⇛(1/Cos θ) + (Sin θ/ Cos θ)

⇛(1+Sin θ)/Cos θ

We know that

Sin² A + Cos² A = 1

⇛Cos² A = 1-Sin² A

⇛Cos A =√(1-Sin² A)

LHS = (1+Sin θ)/√(1- Sin² θ)

⇛ LHS = {1+(a/b)}/√{1-(a/b)²}

= {(b+a)/b}/√(1-(a²/b²))

= {(b+a)/b}/√{(b²-a²)/b²}

= {(b+a)/b}/√{(b²-a²)/b}

= (b+a)/√(b²-a²)

= √{(b+a)(b+a)/(b²-a²)}

⇛ LHS = √{(b+a)(b+a)/(b+a)(b-a)}

Now, (x+y)(x-y) = x²-y²

Where ,

  • x = b and
  • y = a

On cancelling (b+a) then

⇛LHS = √{(b+a)/(b-a)}

⇛RHS

⇛ LHS = RHS

Sec θ + Tan θ = √{(b+a)/(b-a)}

Hence, Proved.

Answer: If Sinθ=a/b then Secθ+Tanθ=√{(b+a)/(b-a)}.

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