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Q.

cos⁡3θcos3⁡θ+sin⁡3θsin3⁡θ is equal to

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a

3cos⁡2θcosec⁡2θ

b

3cot⁡2θcosec⁡2θ

c

12cot⁡2θcosec⁡2θ

d

12tan⁡2θsec⁡2θ

answer is C.

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Detailed Solution

cos⁡3θcos3⁡θ+sin⁡3θsin3⁡θ=4cos3⁡θ−3cos⁡θcos3⁡θ+3sin⁡θ−4sin3⁡θsin3⁡θ=4×3cos2⁡θ−sin2⁡θ4sin2⁡θcos2⁡θ=12cos⁡2θsin⁡2θ×1sin⁡2θ
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