Question

sin30-sin90+2cos0/tan30 tan60
sin30=1/2
sin90=1
cos0=1
tan30=1/√2
tan60=√3​

Answer 1

Answer:

3/2

Step-by-step explanation:

As per the information provided in the question, We have:

• sin30 - sin90 + 2cos0/tan30 tan60

We are asked to find the value of it.

In order to find the value of sin30 - sin90 + 2cos0/tan30 tan60. We need to put the values and simplify it.

sin30 - sin90 + 2cos0/tan30 tan60.

Where,

• sin30° = 1/2
• sin90° = 1
• cos0° = 1
• tan30° = 3/√3
• tan60° = √3

By substituting the values.

$\longmapsto \rm \dfrac{\dfrac{1}{2} - 1 + 2(1)}{\dfrac{\sqrt{3}}{3} \sqrt{3}}$

On simplifying,

$\longmapsto \rm \dfrac{ - \dfrac{1}{2} + 2}{\dfrac{3}{3}}$

$\longmapsto \rm \dfrac{ - \dfrac{1}{2} + 2}{1}$

$\longmapsto \rm \dfrac{ \dfrac{ - 1 + 4}{2} }{1}$

$\longmapsto \rm \dfrac{ \dfrac{ 3}{2} }{1}$

$\longmapsto \rm \dfrac{ 3}{2}$

∴ sin30 - sin90 + 2cos0/tan30 tan60 = 3/2

More to know :

$\bullet\:\sf Trigonometric\:Values :\\\\\boxed{\begin{tabular}{c|c|c|c|c|c}Radians/Angle & 0 & 30 & 45 & 60 & 90\\\cline{1-6}Sin \theta & 0 & \dfrac{1}{2} & \dfrac{1}{\sqrt{2}} & \dfrac{\sqrt{3}}{2} & 1\\\cline{1-6}Cos \theta & 1 & \dfrac{\sqrt{3}}{2}& \dfrac{1}{\sqrt{2}}& \dfrac{1}{2}&0\\\cline{1-6}Tan \theta&0& \dfrac{1}{\sqrt{3}}&1&\sqrt{3}&Not D{e}fined\end{tabular}}$

Answer 2

Question :

sin30-sin90+2cos0/tan30 tan60

Formula Used :

sin30=1/2

sin90=1

cos0=1

tan30=1/√3

tan60=√3

Solution :

$\sf \looparrowright \frac{sin \: 30 \degree - sin \: 90 \degree + 2cos \: 0 \degree}{ \tan \:30 \degree \times tan \: 60 \degree}$

Putting the values given in the formula we get

$\sf \looparrowright \frac{ \frac{1}{2} - 1 + 2 \times 1}{ \frac{1}{ \sqrt{3}} \times \sqrt{3} }$

$\sf \looparrowright \frac{ \frac{1 - 2}{2} +2 }{ \frac{ \sqrt{3} }{ \sqrt{3} } }$

$\sf \looparrowright \frac{ \frac{ - 1 + 4}{2} }{ \frac{ \sqrt{3} }{ \sqrt{3} } }$

$\sf \looparrowright \frac{ \frac{3}{2} }{1}$

$\sf \looparrowright \frac{3}{2}$

So the value is 3/2

KNOW MORE :

Square Relations :

• sin² θ + cos² θ = 1

• sec² θ – tan² θ = 1

• cosec² θ – cot² θ= 1

Quotient Relations :

• sin θ× cosec θ = 1

• cos θ × sec θ = 1

• tan θ × cot θ = 1

Basic :

• sin ∅ = P/H

• cos ∅ = B/H

• tan ∅ = P/B

• cot = B/P

• sec = H/B

• cosec = H/P

Here,

P refers Perpendicular or Height

B refers Base

H refers Hypotentuse

Trigonometric value of standard angles,

$\begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered}\sf Trigonometry\: Table \\ \begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered}\boxed{\boxed{\begin{array}{ |c |c|c|c|c|c|} \bf\angle A & \bf{0}^{ \circ} & \bf{30}^{ \circ} & \bf{45}^{ \circ} & \bf{60}^{ \circ} & \bf{90}^{ \circ} \\ \\ \rm sin A & 0 & \dfrac{1}{2}& \dfrac{1}{ \sqrt{2} } & \dfrac{ \sqrt{3}}{2} &1 \\ \\ \rm cos \: A & 1 & \dfrac{ \sqrt{3} }{2}& \dfrac{1}{ \sqrt{2} } & \dfrac{1}{2} &0 \\ \\ \rm tan A & 0 & \dfrac{1}{ \sqrt{3} }&1 & \sqrt{3} & \rm \infty \\ \\ \rm cosec A & \rm \infty & 2& \sqrt{2} & \dfrac{2}{ \sqrt{3} } &1 \\ \\ \rm sec A & 1 & \dfrac{2}{ \sqrt{3} }& \sqrt{2} & 2 & \rm \infty \\ \\ \rm cot A & \rm \infty & \sqrt{3} & 1 & \dfrac{1}{ \sqrt{3} } & 0\end{array}}}\end{gathered}\end{gathered}\end{gathered} \end{gathered}\end{gathered}\end{gathered}\end{gathered}\end{gathered}\end{gathered}$

Regards

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