MHT CET · Maths · Trigonometric Ratios & Identities
\(\sin 690^{\circ} \times \sec 240^{\circ}=\)
- A \(1\)
- B \(-1\)
- C \(\frac{-1}{2}\)
- D \(\frac{1}{2}\)
Answer & Solution
Correct Answer
(A) \(1\)
Step-by-step Solution
Detailed explanation
\(\sin 690^{\circ} \times \sec 240^{\circ}\)
\(=\sin \left(1 \times 360^{\circ}+330^{\circ}\right) \times \sec \left(180^{\circ}+60^{\circ}\right)\)
\(=\sin 330^{\circ} \times\left(-\sec 60^{\circ}\right)\)
\(=\sin \left(2 \pi-30^{\circ}\right) \times(-2)\)
\(=-2\left(-\sin 30^{\circ}\right)=(-2)\left(-\frac{1}{2}\right)=1\)
\(=\sin \left(1 \times 360^{\circ}+330^{\circ}\right) \times \sec \left(180^{\circ}+60^{\circ}\right)\)
\(=\sin 330^{\circ} \times\left(-\sec 60^{\circ}\right)\)
\(=\sin \left(2 \pi-30^{\circ}\right) \times(-2)\)
\(=-2\left(-\sin 30^{\circ}\right)=(-2)\left(-\frac{1}{2}\right)=1\)
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