Prove that- sin 2 - 6 -cos 2 - 3 -tan 2 - 4 - 1 2-

We know that sin ^ 2 π #160; 6 #160; = 1 4 cos ^ 2 π #160; 3 #160; = 1 4 tan ^ 2 π #160; 4 #160; =1 ⇒ #160; si ...

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Substitution Method to Remove Indeterminate Form

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Question

Prove that: ${sin}^{2} (\frac{π}{6}) + {cos}^{2} (\frac{π}{3}) - {tan}^{2} (\frac{π}{4}) = - \frac{1}{2}$ .

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\frac{{tan}^{2} {(\frac{π}{6})}^{c} + {sin}^{2} {(\frac{π}{6})}^{c} + {cos}^{2} {(\frac{π}{3})}^{c}}{{sec}^{2} {(\frac{π}{4})}^{c} - {cos}^{2} π^{c}} = \frac{1}{\sqrt{3}} sec {(\frac{π}{6})}^{c} + \frac{1}{3} cos {(\frac{π}{3})}^{c}

{sin}^{2} \frac{π}{4} + {cos}^{2} \frac{π}{4} - {tan}^{2} \frac{π}{3}

T h e v a l u e o f {cos}^{2} \frac{π}{4} + {cos}^{2} \frac{π}{6} + {sin}^{2} \frac{π}{6} + {sin}^{2} \frac{π}{3}

{sin}^{2} (π / 8 - A / 2) - {sin}^{2} (π / 8 - A / 2) = \frac{1}{\sqrt{2}} sin A

{cos}^{2} (x) + {cos}^{2} (x + \frac{π}{3}) + {cos}^{2} (x - \frac{π}{3}) = \frac{3}{2}

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