TS EAMCET · Maths · Trigonometric Equations
If \(\sin x+\sin y=p, \cos x+\cos y=q\), then \(\sec (x+y)=\)
- A \(\frac{2 p q}{p^2+q^2}\)
- B \(\frac{p^2+q^2}{q^2-p^2}\)
- C \(\frac{2 p q}{\sqrt{p^2+q^2}}\)
- D \(\frac{p+q}{p^2+q^2}\)
Answer & Solution
Correct Answer
(B) \(\frac{p^2+q^2}{q^2-p^2}\)
Step-by-step Solution
Detailed explanation
It is given that, \(\sin x+\sin y=p\), and \(\cos x+\cos y=q\) So, on squaring and adding, we get \(2+2 \cos (x-y)=p^2+q^2\) On squaring and subtracting, we get \(\cos 2 x+\cos 2 y+2 \cos (x+y)=q^2-p^2\) \(\Rightarrow 2 \cos (x+y)[\cos (x-y)+1]=q^2-p^2\) From Eqs. (i) and (ii),…
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