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ML Aggarwal Solutions Class 9 Mathematics Solutions for Trigonometric Ratios Exercise 17 in Chapter 17 - Trigonometric Ratios
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Given sin = p/q find sin + cos
Answer:
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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