Question

tan² 60°+4 cos² 45°+ 3cosec² 60°+ 2 cos² 90°/ 2 cosec 30° + 3 sec 60° - 7/3 cot² 30°



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Answer 1

ANSWER:

• The value of the given trigonometric ratio is 3.

TRIGONOMETRIC RATIOS:

$\large\displaystyle\begin{tabular}{|c|c|c|c|c|c|}\cline{1-6}&0^{\circ}&30^{\circ}&45^{\circ}&60^{\circ}&90^{\circ}\cline{1-6}\cline{1-6}Sin&0& \dfrac{1}{2} & \dfrac{1}{\sqrt{2}} & \dfrac{\sqrt{3}}{2} &1\cline{1-6}\cline{1-6}Cos&1& \dfrac{\sqrt{3}}{2} & \dfrac{1}{\sqrt{2}} & \dfrac{1}{2} &0\cline{1-6}\cline{1-6}Tan&0& \dfrac{1}{\sqrt{3}} &1& \sqrt{3} &\infty\cline{1-6}\cline{1-6}\end{tabular}$

$\large\displaystyle\begin{tabular}{|c|c|c|c|c|c|}\cline{1-6}&0^{\circ}&30^{\circ}&45^{\circ}&60^{\circ}&90^{\circ}\cline{1-6}\cline{1-6}cot&\infty& \sqrt{3} &1& \dfrac{1}{\sqrt{3}} &0\cline{1-6}\cline{1-6}Sec&1& \dfrac{2}{\sqrt{3}} & \sqrt{2} &2&\infty\cline{1-6}\cline{1-6}Cosec&\infty&2&\sqrt{2}& \dfrac{2}{\sqrt{3}} &1\cline{1-6}\cline{1-6}\end{tabular}$

EXPLANATION:

From the above table

•°• tan 60° = √3

°•° tan² 60° = 3

•°• cos 45° = 1/√2

°•° 4 cos² 45° = 4 × 1/2 = 2

•°• cosec 60° = 2/√3

°•° 3 cosec² 60° = 3 × 4/3 = 4

•°• cos 90° = 0

°•° 2 cos² 90° = 0

•°• cosec 30° = 2

°•° 2 cosec 30° = 4

•°• sec 60° = 2

°•° 3 sec 30° = 6

•°• cot 30° = √3

°•° 7/3 cot² 30° = 7/3 × 3 = 7

We want to find the value of:

$\dfrac{tan^2 60^{\circ}+4 cos^2 45^{\circ}+ 3cosec^2 60^{\circ}+ 2 cos^2 90^{\circ}}{2 cosec 30^{\circ} + 3 sec 60^{\circ} - 7/3 cot^2 30^{\circ}}$

Substitute the respective values

$\dfrac{3+2+4 +0}{4+ 6- 7}$

=> 9/3 = 3

${\dfrac{tan^2 60^{\circ}+4 cos^2 45^{\circ}+ 3cosec^2 60^{\circ}+ 2 cos^2 90^{\circ}}{2 cosec 30^{\circ} + 3 sec 60^{\circ} - 7/3 cot^2 30^{\circ}} = 3 }$

Answer 2

Answer:

3

Step-by-step explanation:

tan 60° = √3

cos 45° = 1/√2

cosec 60° = 2/√3

cos 90° = 0

cosec 30° = 2

sec 60° = 2

cot 30° = √3

Substitute the values properly. The image above shows the steps involved.

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