Question

This is a challenge

Answer 1

Answer: $\dfrac{-21}{12}$

Step-by-step explanation:

$\text{The given expression is } A = \dfrac{3}{4}\cot^230^{\circ} + 3\sin^260^{\circ} - 2 \csc^260^{\circ} - \dfrac{3}{4}\tan^230^{\circ}\\\text{Using trigonometric tables,} \\\cot 30^{\circ} = \sqrt{3} \\\sin60^{\circ} = \dfrac{\sqrt{3}}{2}\\\csc60^{\circ} = \dfrac{1}{\sin60^{\circ}} = \dfrac{2}{\sqrt{3}}\\\tan30^{\circ} = \dfrac{1}{\cot30^{\circ}} = \dfrac{1}{\sqrt{3}}\\\\\text{Substituting these in the given expression, }$

$A = \dfrac{3}{4}(\sqrt{3})^2 + 3(\dfrac{\sqrt{3}}{2})^2 - 2(\dfrac{2}{\sqrt{3}})^2 - \dfrac{3}{4}(\dfrac{1}{\sqrt{3}})^2$

$A = \dfrac{9}{4} + \dfrac{9}{4} - \dfrac{8}{3} - \dfrac{1}{4}$

$A = \dfrac{-21}{12}$

Answer 2

Qᴜᴇsᴛɪᴏɴ :-

find the value of :- (3/4)cot²30° + 3sin²60° - 2cosec²60° - (3/4)tan²30° ?

Value used :-

• cot30° = √3
• sin60° = (√3/2)
• cosec60° = (2/√3)
• tan30° = (1/√3)

Sᴏʟᴜᴛɪᴏɴ :-

→ (3/4)cot²30° + 3sin²60° - 2cosec²60° - (3/4)tan²30°

→ (3/4)cot²30 - (3/4)tan²30° + 3sin²60° - 2cosec²60°

→ (3/4)[cot²30 - tan²30°] + 3sin²60° - 2cosec²60°

Putting values Now, we get,

→ (3/4)[(√3)² - (1/√3)²] + 3*(√3/2)² - 2(2/√3)²

→ (3/4)[ 3 - 1/3 ] + 3*(3/4) - 2*(4/3)

→ (3/4)*(8/3) + (9/4) - (8/3)

→ 2 + (9/4) - (8/3)

Taking LCM of Denominators now,

→ (24 + 27 - 32) / 12

→ (51 - 32)/12

→ (19/12) = 1(7/12) (Ans.)