Find he value of- 3 cosec2 60- - 2 2 30- - 2 45-

3 cosec^2 60^∘ 2 ^2 30^∘ + ^2 45^∘ = 0 L.H.S. = #160; 3 cosec^2 60^∘ 2 ^2 30^∘ + ^2 45^∘ = 3 (2√(3))^2 2 (√(3))^2 + (√(2))^2 = 3 (43) ...

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Standard X

Mathematics

Range of Trigonometric Ratios from 0 to 90 Degrees

Find he value...

Question

Find he value of: $3 {cosec}^{2} 60^{\circ} - 2 {cot}^{2} 30^{\circ} + {sec}^{2} 45^{\circ}$

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Solution

$3 {cosec}^{2} 60^{\circ} - 2 {cot}^{2} 30^{\circ} + {sec}^{2} 45^{\circ} = 0$
L.H.S. = $3 {cosec}^{2} 60^{\circ} - 2 {cot}^{2} 30^{\circ} + {sec}^{2} 45^{\circ}$
= $3 (\frac{2}{\sqrt{3}})^{2} - 2 (\sqrt{3})^{2} + (\sqrt{2})^{2}$
= $3 (\frac{4}{3}) - 2 (3) + 2$
= $4 - 6 + 2$
= $0$

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Similar questions

Find the value of $\frac{4}{3} t a n^{2} 60^{\circ} + 3 c o s^{2} 30^{\circ} - 2 s e c^{2} 30^{\circ} - \frac{3}{4} c o t^{2} 60^{\circ}$

The value of
$\frac{c o t^{2} 30^{\circ} + 8 s i n^{2} 45^{\circ} + \frac{3}{2} s e c^{2} 30^{\circ} + 2 c o s^{2} 90^{\circ}}{2 s e c 60^{\circ} + 3 c o s e c 30^{\circ} - \frac{7}{3} t a n^{2} 60^{\circ}}$ is :

Q. (3cos²60° + 2cot²30° − 5sin²45°) = ?
(a)

\frac{13}{6}

(b)

\frac{17}{4}

$3 c o s^{2} 60^{o} + 2 c o t^{2} 30^{o} - 5 s i n^{2} 45^{o}$ = ?

(a) $\frac{13}{6}$ (a) $\frac{17}{4}$ (c) 1 (d) 4

Q. (3cos²60° + 2cot²30° – 5sin²45°) = ?
(a) 1

(b) 4

(c)

\frac{17}{4}

(d)

\frac{13}{6}

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Find he value of: 3cosec260∘−2cot230∘+sec245∘

3cosec260∘−2cot230∘+sec245∘=0L.H.S. = 3cosec260∘−2cot230∘+sec245∘= 3(2√3)2−2(√3)2+(√2)2= 3(43)−2(3)+2= 4−6+2= 0

Find he value of: $3 {cosec}^{2} 60^{\circ} - 2 {cot}^{2} 30^{\circ} + {sec}^{2} 45^{\circ}$

$3 {cosec}^{2} 60^{\circ} - 2 {cot}^{2} 30^{\circ} + {sec}^{2} 45^{\circ} = 0$
L.H.S. = $3 {cosec}^{2} 60^{\circ} - 2 {cot}^{2} 30^{\circ} + {sec}^{2} 45^{\circ}$
= $3 (\frac{2}{\sqrt{3}})^{2} - 2 (\sqrt{3})^{2} + (\sqrt{2})^{2}$
= $3 (\frac{4}{3}) - 2 (3) + 2$
= $4 - 6 + 2$
= $0$