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Question 149 of 175

Q.(a) If A+B+C=πA+B+C=\pi, prove that cos⁡2A+cos⁡2B+cos⁡2C=1−2cos⁡Acos⁡Bcos⁡C\cos^2 A + \cos^2 B + \cos^2 C = 1 - 2\cos A \cos B \cos C. OR

(b) If log⁡xy−z=log⁡yz−x=log⁡zx−y\dfrac{\log x}{y-z} = \dfrac{\log y}{z-x} = \dfrac{\log z}{x-y} then prove that xyz=1xyz = 1
Puducherry TnboardTamil Nadu HSC First Year (DGE) Board 2020Subjective· 5mImportance★★★★★
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Using C = π − (A+B) and the identity cos⁡2A+cos⁡2B=1+cos⁡(A+B)cos⁡(A−B)\cos^2A+\cos^2B = 1+\cos(A+B)\cos(A-B), the given identity follows directly.

Given A+B+C=πA+B+C=\pi, so C=π−(A+B)C = \pi-(A+B), which gives cos⁡C=−cos⁡(A+B)\cos C = -\cos(A+B).

Step 1: Write cos⁡2A+cos⁡2B\cos^2A+\cos^2B using the double-angle form:

cos⁡2A+cos⁡2B=1+cos⁡2A2+1+cos⁡2B2=1+cos⁡(A+B)cos⁡(A−B)\cos^2A+\cos^2B = \dfrac{1+\cos2A}{2}+\dfrac{1+\cos2B}{2} = 1+\cos(A+B)\cos(A-B)

Step 2: Add cos⁡2C\cos^2C:

cos⁡2A+cos⁡2B+cos⁡2C=1+cos⁡(A+B)cos⁡(A−B)+cos⁡2C\cos^2A+\cos^2B+\cos^2C = 1+\cos(A+B)\cos(A-B)+\cos^2C

Since cos⁡(A+B)=−cos⁡C\cos(A+B) = -\cos C:

=1−cos⁡Ccos⁡(A−B)+cos⁡2C=1+cos⁡C[cos⁡C−cos⁡(A−B)]= 1-\cos C\cos(A-B)+\cos^2C = 1+\cos C[\cos C-\cos(A-B)]

Step 3: Expand cos⁡C\cos C using C=π−(A+B)C=\pi-(A+B):

cos⁡C=−cos⁡(A+B)=−cos⁡Acos⁡B+sin⁡Asin⁡B\cos C = -\cos(A+B) = -\cos A\cos B+\sin A\sin B

So: …

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