# ML Aggarwal Solutions Class 9 Mathematics Solutions for Trigonometric Ratios Exercise 17 in Chapter 17 - Trigonometric Ratios

Question 9 Trigonometric Ratios Exercise 17

Given sin = p/q find sin + cos

Given that sin θ = p/q

Which implies,

AB/AC = p/q

Let AB = px

And then AC = qx

In right-angled triangle ABC

By Pythagoras theorem,

We get

\begin{array}{l} A C^{2}=A B^{2}+B C^{2} \\ B C^{2}=A C^{2}-A B^{2} \\ B C^{2}=q^{2} x^{2}-p^{2} x^{2} \\ B C^{2}=\left(q^{2}-p^{2}\right) x^{2} \\ B C=V\left(q^{2}-p^{2}\right) x \end{array}

In right angled triangle ABC,

Cos θ = base/ hypotenuse

= BC/AC

\begin{aligned} &=v\left(q^{2}-p^{2}\right) x / q x\\ &=v\left(q^{2}-p^{2}\right) / q\\ &\text { Now, }\\ &\sin \theta+\cos \theta=p / q+v\left(q^{2}-p^{2}\right) / q\\ &=\left[p+V\left(q^{2}-p^{2}\right)\right] / q \end{aligned}

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