Question

if sin theta = p/q then value of tan theta+sec theta is?
Identity answers only

Answer 1

Answer: Check the picture

Answer 2

Given,

Sinθ = $\frac{p}{q}$

To find,

Tanθ + Secθ

Solution,

From the given value of Sinθ, we have

The hypotenuse of the triangle = q.

The height of the triangle = p.

By using Pythagoras theorem,

       $Base^{2}$ + $p^{2} = q^{2}$

        $Base^{2}$ $= q^{2} - p^{2}$

       $Base = \sqrt{ q^{2} - p^{2} }$

Now, using the formula

$($Secθ$)^{2}$ - $($Tanθ$)^{2}$ = 1

This can be further factorized to,

(Sexθ - Tanθ)(Secθ + Tanθ) = 1  --------eq(1)

Secθ = 1/Cosθ and Tanθ = Sinθ/Cosθ

Substituting the value in eq(1)

(1/Cosθ - Sinθ/Cosθ)(Secθ + Tanθ) = 1

Calculating the value of (1/Cosθ - Sinθ/Cosθ),

$= \frac{q}{\sqrt{ q^{2} - p^{2} }} - \frac{p}{\sqrt{ q^{2} - p^{2} }} \\= \frac{p - q}{\sqrt{ q^{2} - p^{2} }} }$

Substituting the value in the equation above, we get the required value of Tanθ + Secθ.

Tanθ + Secθ = $\frac{\sqrt{q^{2} - p^{2} }}{ q - p}$

Therefore, value of Tanθ + Secθ is $\frac{\sqrt{q^{2} - p^{2} }}{ q - p}$.