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Questions  >>  CBSE XI  >>  Math  >>  Trigonometric Functions
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Q)

$2\sin ^2\large\frac{3\pi}{4}$$+2\cos^2\large\frac{\pi}{4}$$+2\sec^2\large\frac{\pi}{3}$ =

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A)
2sin2⁡(π−π4)" role="presentation" style="position: relative;">2sin2(ππ4)2sin2⁡(π−π4)2\sin^2(\pi-\large\frac{\pi}{4})+2(cosπ4)2" role="presentation" style="position: relative;">+2(cosπ4)2+2(cosπ4)2+2(\cos\large\frac{\pi}{4})^2+2(secπ3)2" role="presentation" style="position: relative;">+2(secπ3)2+2(secπ3)2+2(\sec\large\frac{\pi}{3})^2
⇒2(sinπ4)2" role="presentation" style="position: relative;">2(sinπ4)2⇒2(sinπ4)2\Rightarrow 2(\sin\large\frac{\pi}{4})^2+2×(12)2" role="presentation" style="position: relative;">+2×(12)2+2×(12)2+2\times (\large\frac{1}{\sqrt 2})^2+2×(2)2" role="presentation" style="position: relative;">+2×(2)2+2×(2)2+2\times (2)^2
⇒2×(12)2" role="presentation" style="position: relative;">2×(12)2⇒2×(12)2\Rightarrow 2\times (\large\frac{1}{\sqrt 2})^2+1+8=2×12" role="presentation" style="position: relative;">+1+8=2×12+1+8=2×12+1+8=2\times \large\frac{1}{2}+1+8" role="presentation" style="position: relative;">+1+8+1+8+1+8
$1+1+8=10⇒1+1+8=10\Rightarrow 1+1+8=10$
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