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JEE MATHEMATICS MCQ: TRIGONOMETRY RATIOS

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Application +4 | -1
In a Δ   ABC,   ∠  A=π2\Delta \,\;ABC,\;\,\angle \;A = \frac{\pi }{2}, then cos⁡2  B+cos⁡2  C{\cos ^2}\,\,B + {\cos ^2}\,\,C equals

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Correct option: C
1\mathrm{1}
Option C (1) is correct because in a right-angled triangle, B+C=π2B + C = \frac{\pi}{2}, and thus cos⁡2 B+cos⁡2 C=cos⁡2 B+sin⁡2 B=1{\cos ^2}\,B + {\cos ^2}\,C = {\cos ^2}\,B + {\sin ^2}\,B = 1.

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