sin ^{2} 30^{circ} cos ^{2} 45^{circ}+5 tan ^ | vedclass

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Question

(A) આપેલ પદાવલિ: $\sin ^{2} 30^{\circ} \cos ^{2} 45^{\circ}+5 	an ^{2} 30^{\circ}+\frac{3}{2} \sin ^{2} 90^{\circ}-3 \cos ^{2} 90^{\circ}$ત્રિકોણમિતી

Vedclass · Accepted Answer

$3 \frac{7}{24}$

Vedclass · Answer

(A) આપેલ પદાવલિ: $\sin ^{2} 30^{\circ} \cos ^{2} 45^{\circ}+5 	an ^{2} 30^{\circ}+\frac{3}{2} \sin ^{2} 90^{\circ}-3 \cos ^{2} 90^{\circ}$ત્રિકોણમિતીય કિંમતો મૂકતા: $\sin 30^{\circ} = \frac{1}{2}$,$\cos 45^{\circ} = \frac{1}{\sqrt{2}}$,$	an 30^{\circ} = \frac{1}{\sqrt{3}}$,$\sin 90^{\circ} = 1$,$\cos 90^{\circ} = 0$.$= \left[\left(\frac{1}{2}ight)^{2} 	imes \left(\frac{1}{\sqrt{2}}ight)^{2}ight] + 5\left(\frac{1}{\sqrt{3}}ight)^{2} + \frac{3}{2}(1)^{2} - 3(0)^{2}$$= \left(\frac{1}{4} 	imes \frac{1}{2}ight) + 5\left(\frac{1}{3}ight) + \frac{3}{2} - 0$$= \frac{1}{8} + \frac{5}{3} + \frac{3}{2}$$8, 3, 2$ નો લઘુત્તમ સામાન્ય અવયવી $(LCM)$ $24$ લેતા:$= \frac{3 + 40 + 36}{24} = \frac{79}{24}$મિશ્ર અપૂર્ણાંકમાં ફેરવતા: $\frac{79}{24} = 3 \frac{7}{24}$.

$\sin ^{2} 30^{\circ} \cos ^{2} 45^{\circ}+5 \tan ^{2} 30^{\circ}+\frac{3}{2}\sin ^{2} 90^{\circ}-3 \cos ^{2} 90^{\circ}$ ની કિંમત શોધો.

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જો $A + B + C = \pi$ હોય,તો $\tan^2 \frac{A}{2} + \tan^2 \frac{B}{2} + \tan^2 \frac{C}{2}$ હંમેશા કેટલા થાય?

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