Cos70-sin20-cos59-sin31-8 sin230- is equal to

Cos70^∘/sin20^∘+cos59^∘/sin31^∘ 8 sin^2 30^∘ ⇒sin20^∘/sin20^∘ + nbsp;sin31^∘/sin31^∘ 8(1/2)^2 (∵ cosθ = sin(90^∘ θ), sin30^∘ = (1/2) ⇒ 1 + 1 8 nbsp; ...

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Byju's Answer

Standard IX

Mathematics

Value of Standard Angle(30,60,90)

cos70∘/sin20∘...

Question

$\frac{c o s 70^{\circ}}{s i n 20^{\circ}} + \frac{c o s 59^{\circ}}{s i n 31^{\circ}} - 8 s i n^{2} 30^{\circ}$ is equal to
0

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Solution

The correct option is A 0
$\frac{c o s 70^{\circ}}{s i n 20^{\circ}} + \frac{c o s 59^{\circ}}{s i n 31^{\circ}} - 8 s i n^{2} 30^{\circ} \Rightarrow \frac{s i n 20^{\circ}}{s i n 20^{\circ}} + \frac{s i n 31^{\circ}}{s i n 31^{\circ}} - 8 (\frac{1}{2})^{2} (∵ c o s θ = s i n (90^{\circ} - θ), s i n 30^{\circ} = (\frac{1}{2}) \Rightarrow 1 + 1 - 8 \times \frac{1}{4} \Rightarrow 2 - 2 = 0$

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\frac{c o s 70^{\circ}}{s i n 20^{\circ}} + \frac{c o s 59^{\circ}}{s i n 31^{\circ}} - 8 s i n^{2} 30^{\circ}

is equal to

Q. Find:

\frac{c o s 70^{\circ}}{s i n 20^{\circ}} + \frac{c o s 59^{\circ}}{s i n 31^{\circ}} - 8 s i n^{2} 30^{\circ} =

______.

Q. Evaluate

\frac{c o s 70^{\circ}}{s i n 20^{\circ}} + \frac{c o s 59^{\circ}}{s i n 31^{\circ}} - 8 s i n^{2} 30^{\circ} = 0

$\frac{c o s 70^{\circ}}{s i n 20^{\circ}} + \frac{c o s 59^{\circ}}{s i n 31^{\circ}} - 8 s i n^{2} 30^{\circ} =$

Q. Find the value of the following expression.

\frac{c o s 70^{\circ}}{s i n 20^{\circ}} + \frac{c o s 59^{\circ}}{s i n 31^{\circ}} - 8 s i n^{2} 30^{\circ}

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cos70∘sin20∘+cos59∘sin31∘−8sin230∘ is equal to0

The correct option is A 0cos70∘sin20∘+cos59∘sin31∘−8sin230∘⇒sin20∘sin20∘+sin31∘sin31∘−8(12)2(∵ cosθ=sin(90∘−θ), sin30∘=(12)⇒1+1−8×14⇒2−2=0

$\frac{c o s 70^{\circ}}{s i n 20^{\circ}} + \frac{c o s 59^{\circ}}{s i n 31^{\circ}} - 8 s i n^{2} 30^{\circ}$ is equal to
0