2 tan 2 45-3 cos 2 30-sin 2 60-

Tan 45^∘ = 1, cos 30^∘ = √(3)/2 and sin 60^∘ = √(3)/2 ⇒ 2 tan^245^∘+3cos^230^∘ sin^260^∘ = 2(1)^2 + 3 (√(3)/2)^2 (√(3)/2)^2 = 2 + 9/4 3/4 = 8 + 9 3/4 ...

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Standard X

Mathematics

Complementary Trigonometric Ratios

2 tan 2 45∘+3...

Question

$2 t a n^{2} 45^{\circ} + 3 c o s^{2} 30^{\circ} - s i n^{2} 60^{\circ} =$ ___

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Solution

$t a n 45^{\circ} = 1, c o s 30^{\circ} = \frac{\sqrt{3}}{2} a n d s i n 60^{\circ} = \frac{\sqrt{3}}{2} \Rightarrow 2 t a n^{2} 45^{\circ} + 3 c o s^{2} 30^{\circ} - s i n^{2} 60^{\circ} = 2 (1)^{2} + 3 (\frac{\sqrt{3}}{2})^{2} - (\frac{\sqrt{3}}{2})^{2} = 2 + \frac{9}{4} - \frac{3}{4} = \frac{8 + 9 - 3}{4} = \frac{14}{4} = \frac{7}{2}$

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Similar questions

$2 t a n^{2} 45^{\circ} + 3 c o s^{2} 30^{\circ} - s i n^{2} 60^{\circ} =$ ___

Q. Evaluate:

2 {tan}^{2} 45 + {cos}^{2} 30^{\circ} - {sin}^{2} 60^{\circ}

Q. Question 1 (ii)
Evaluate the following:
(ii)

2 t a n^{2} 45^{\circ} + c o s^{2} 30^{\circ} - s i n^{2} 60^{\circ}

Q. Question 1 (ii)
Evaluate the following:
(ii)

2 t a n^{2} 45^{\circ} + c o s^{2} 30^{\circ} - s i n^{2} 60^{\circ}

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(i) 2 tan ^₂ 45

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sin ^₂ 60

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+ cosec ^₂30

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+cot 90

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2tan245∘+3cos230∘−sin260∘=___

tan45∘=1, cos30∘=√32 and sin60∘=√32⇒2tan245∘+3cos230∘−sin260∘=2(1)2+3(√32)2−(√32)2=2+94−34=8+9−34=144=72

$2 t a n^{2} 45^{\circ} + 3 c o s^{2} 30^{\circ} - s i n^{2} 60^{\circ} =$ ___