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निम्नलिखित समीकरणों को हल कीजिए-1. sine + cos 0 = 12. |
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Answer» Squaring first equatiosin^2(a)+cos^2(a) +2sin( a) cos( a)=21+2sin(a)cos(a)=22sin(a)cos(a)=1sin(a)cos(a)=1/2tan a+cot a=sin a/cos a +cos a/sin a=sin ^2 a +cos ^2 a/sin a cos a=1/(1/2)=2 sin²o + cos²o = 22 the answer is right ( sinx + cosx)^2=( V2)^2; ( sinx^2 + cosx^2 + 2sinxcosx) = 2; 2 sinxcosx = 2-1=1; sinxcosx = 1/2 sin θ + cos θ = √2 Squaring on both sides,(sin θ + cos θ)² = √2² sin²θ + cos²θ + 2sinθ.cosθ = 2 1 + 2sinθ.cosθ = 2 2sinθ.cosθ = 2-1 2sinθ.cosθ = 1 sinθ.cosθ = ½ sinθ.cosθ = (1/√2) × (1/√2) sinθ = cos θ = 1/√2 θ = 45* |
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