MHT CET · Maths · Trigonometric Equations
If \(\sec x=\frac{25}{24}\) and \(x\) lies in first quadrant, then \(\sin \frac{x}{2}+\cos \frac{x}{2}=\)
- A \(\frac{6}{5 \sqrt{2}}\)
- B \(\frac{8}{5 \sqrt{2}}\)
- C \(\frac{7}{5 \sqrt{2}}\)
- D \(\frac{1}{5 \sqrt{2}}\)
Answer & Solution
Correct Answer
(B) \(\frac{8}{5 \sqrt{2}}\)
Step-by-step Solution
Detailed explanation
We have \(\cos x=\frac{24}{25} \Rightarrow \sin x=\frac{7}{25} \ldots\left[\because x\right.\) lies in \(1^{\text {st }}\) quadrant \(]\) Also \(\left(\sin \frac{x}{2}+\cos \frac{x}{2}\right)^2=1+\sin x=1+\frac{7}{25}=\frac{32}{25}\)
\(\therefore \sin \frac{x}{2}+\cos \frac{x}{2}=\sqrt{\frac{32}{25}}=\frac{4 \sqrt{2}}{5}=\frac{8}{5 \sqrt{2}}\)
\(\therefore \sin \frac{x}{2}+\cos \frac{x}{2}=\sqrt{\frac{32}{25}}=\frac{4 \sqrt{2}}{5}=\frac{8}{5 \sqrt{2}}\)
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