Solve- 2 sin2 30- - 3 cos2 60- - cot2 30- -2 MARKS-

Trigonometric values: 1 Mark Application: 1 Mark 2 sin^2 30^∘ 3 cos^2 60^∘ + cot^2 30^∘ 2 ( 1/2 )^2 3 ( 1/2 )^2 + (√(3))^2 =2 ( 1/4 ) 3 ( 1/4 )+3 =1/2 ...

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Byju's Answer

Standard X

Mathematics

Range of Trigonometric Ratios from 0 to 90 Degrees

Solve: 2 sin2...

Question

Solve:
$2 s i n^{2} 30^{\circ} - 3 c o s^{2} 60^{\circ} + c o t^{2} 30^{\circ}$ [2 MARKS]

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Solution

Trigonometric values: 1 Mark
Application: 1 Mark

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

$2 {(\frac{1}{2})}^{2} - 3 {(\frac{1}{2})}^{2} + (\sqrt{3})^{2}$

$= 2 (\frac{1}{4}) - 3 (\frac{1}{4}) + 3$

$= \frac{1}{2} - \frac{3}{4} + 3$

$= \frac{2 - 3 + 12}{4}$

$= \frac{11}{4}$

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Similar questions

s i n^{2} 30^{\circ} + c o s^{2} 60 + {tan}^{2} 45^{\circ} + s e c^{2} 60^{\circ} - c o {sec}^{2} 30^{\circ} = \frac{3}{2}

Q. Solve

{csc}^{2} 45 . {sec}^{2} 30 . {sin}^{2} 90 . cos 60 = 1 \frac{1}{3}

Q. Prove that

(i) s e c^{2} θ + c o s e c^{2} θ = s e c^{2} θ c o s e c^{2} θ

(i i) t a n^{2} θ - s i n^{2} θ = t a n^{2} θ s i n^{2} θ

(i i i) t a n^{2} θ + c o t^{2} θ + 2 = s e c^{2} θ c o s e c^{2} θ

[3 MARKS]

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

= ________ .

Q. Solve:

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

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Solve: 2 sin2 30∘−3 cos260∘+cot2 30∘ [2 MARKS]

Trigonometric values: 1 Mark Application: 1 Mark 2 sin2 30∘−3 cos260∘+cot2 30∘ 2(12)2−3(12)2+(√3)2 =2(14)−3(14)+3 =12−34+3 =2−3+124 =114

Solve:
$2 s i n^{2} 30^{\circ} - 3 c o s^{2} 60^{\circ} + c o t^{2} 30^{\circ}$ [2 MARKS]