જો left[1-cos left(frac{pi}{2}+alpharight)+si | vedclass

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(B) આપેલ પદાવલિ: $\left[1-\cos \left(\frac{\pi}{2}+\alpha\right)+\sin \left(\frac{3 \pi}{2}+\alpha\right)\right]^2+\left[1-\sin \left(\frac{3 \pi}{2}-

Vedclass · Accepted Answer

$52$

Vedclass · Answer

(B) આપેલ પદાવલિ: $\left[1-\cos \left(\frac{\pi}{2}+\alpha\right)+\sin \left(\frac{3 \pi}{2}+\alpha\right)\right]^2+\left[1-\sin \left(\frac{3 \pi}{2}-\alpha\right)-\cos \left(\frac{3 \pi}{2}+\alpha\right)\right]^2=a+b \sin ^2\left(\frac{\pi}{4}+\alpha\right)$.ત્રિકોણમિતીય નિત્યસમનો ઉપયોગ કરતા: $\cos(\frac{\pi}{2}+\alpha) = -\sin \alpha$,$\sin(\frac{3\pi}{2}+\alpha) = -\cos \alpha$,$\sin(\frac{3\pi}{2}-\alpha) = -\cos \alpha$,અને $\cos(\frac{3\pi}{2}+\alpha) = \sin \alpha$.પદાવલિ આ મુજબ બનશે: $(1+\sin \alpha-\cos \alpha)^2+(1+\cos \alpha-\sin \alpha)^2 = a+b \sin ^2(\frac{\pi}{4}+\alpha)$.બંને કૌંસનું વિસ્તરણ કરતા: $(1+\sin^2 \alpha+\cos^2 \alpha+2\sin \alpha-2\sin \alpha \cos \alpha-2\cos \alpha) + (1+\cos^2 \alpha+\sin^2 \alpha+2\cos \alpha-2\sin \alpha \cos \alpha-2\sin \alpha) = a+b \sin^2(\frac{\pi}{4}+\alpha)$.સાદુરૂપ આપતા: $4-4\sin \alpha \cos \alpha = a+b(\sin \frac{\pi}{4} \cos \alpha+\cos \frac{\pi}{4} \sin \alpha)^2$.$4-4\sin \alpha \cos \alpha = a+\frac{b}{2}(\cos \alpha+\sin \alpha)^2 = a+\frac{b}{2}(1+2\sin \alpha \cos \alpha) = (a+\frac{b}{2}) + b\sin \alpha \cos \alpha$.સહગુણકોની સરખામણી કરતા: $b = -4$ અને $a+\frac{b}{2} = 4$ $\Rightarrow a-2 = 4$ $\Rightarrow a = 6$.આમ,$a^2+b^2 = 6^2+(-4)^2 = 36+16 = 52$.

જો $\left[1-\cos \left(\frac{\pi}{2}+\alpha\right)+\sin \left(\frac{3 \pi}{2}+\alpha\right)\right]^2+\left[1-\sin \left(\frac{3 \pi}{2}-\alpha\right)-\cos \left(\frac{3 \pi}{2}+\alpha\right)\right]^2=a+b \sin ^2\left(\frac{\pi}{4}+\alpha\right)$,હોય તો $a^2+b^2=$

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