Prove that : tan(60∘ +θ)tan(30∘ −θ)cos2(45∘ +θ)+cos2(45∘ −θ)=1.

We know, 2cos2x=1+cos2x.∴cos2x=21+cos2xcos2(455+θ)=21+cos(90o+θ)=21−sinθcos2(45o−θ)=21+cos(900−θ)=21+sinθtan(60o+θ)=1−tan60otanθtan60o+tanθ=1−3tanθ3+tanθtan(30o−θ)=1+tan30otanθtan30o−tanθ=3+tanθ1−3tanθ∴LHS=1−3tanθ3+tanθ×(1−3 tanθ3+tanθ)−121−sinθ+21+sinθ=21+21=1=RHS (proved)

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Prove that : { { cos }^{ 2 }({ 45 }^{ }+ )+{ cos }^{ 2 }({

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Prove that : $\frac{cos ^{2} ( 45 ^{\circ} + θ ) + cos ^{2} ( 45 ^{\circ} - θ )}{t an ( 60 ^{\circ} + θ ) t an ( 30 ^{\circ} - θ )} = 1.$

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Prove that :

\frac{cos ^{2} ( 45 ^{\circ} + θ ) + cos ^{2} ( 45 ^{\circ} - θ )}{t an ( 60 ^{\circ} + θ ) t an ( 30 ^{\circ} - θ )} = 1.

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Text solutionVerified

We know,

2 cos^{2} x = 1 + cos 2 x

∴ cos^{2} x = \frac{1 + cos ^{2} x}{2}

cos^{2} (4 5^{5} + θ) = \frac{1 + cos ( 9 0 ^{o} + θ )}{2} = \frac{1 - sin θ}{2}

cos^{2} (4 5^{o} - θ) = \frac{1 + cos ( 9 0 ^{0} - θ )}{2} = \frac{1 + sin θ}{2}

tan (6 0^{o} + θ) = \frac{tan 6 0 ^{o} + tan θ}{1 - tan 6 0 ^{o} tan θ} = \frac{3 + tan θ}{1 - 3 tan θ}

tan (3 0^{o} - θ) = \frac{tan 3 0 ^{o} - tan θ}{1 + tan 3 0 ^{o} tan θ} = \frac{1 - 3 tan θ}{3 + tan θ}

∴ L H S = \frac{\frac{1 - sin θ}{2} + \frac{1 + sin θ}{2}}{\frac{3 + tan θ}{1 - 3 tan θ} \times ( \frac{3 + tan θ}{1 - 3 t an θ} ) ^{- 1}} = \frac{1}{2} + \frac{1}{2} = 1 = R H S

(proved)

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Question Text	Prove that : $\frac{cos ^{2} ( 45 ^{\circ} + θ ) + cos ^{2} ( 45 ^{\circ} - θ )}{t an ( 60 ^{\circ} + θ ) t an ( 30 ^{\circ} - θ )} = 1.$
Answer Type	Text solution:1
Upvotes	150

Prove that : tan(60∘ +θ)tan(30∘ −θ)cos2(45∘ +θ)+cos2(45∘ −θ)​=1.

Text solutionVerified

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Prove that : $\frac{cos ^{2} ( 45 ^{\circ} + θ ) + cos ^{2} ( 45 ^{\circ} - θ )}{t an ( 60 ^{\circ} + θ ) t an ( 30 ^{\circ} - θ )} = 1.$