Question

Heya everyone.
Find the value of the following..
$\sf \cos(1^{ \circ} ) \times \cos(2 ^ {\circ}) \times ... \cos(90^{ \circ} )$
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Answer 1

Answer:

the required value is 0

Step-by-step explanation:

0× anything =0

We know cos 90degree =0

So the answer is 0

Answer 2

We have to find out the value of :-

$\rm \longrightarrow \cos(1^{ \circ} ) \times \cos(2 ^ {\circ}) \times ... \cos(90^{ \circ} )$

We know that :-

Value of cos 90° = 0

$\rm \longrightarrow \cos(1^{ \circ} ) \times \cos(2 ^ {\circ}) \times ... \cos(89^{ \circ} )\times 0$

$\rm \longrightarrow 0$

Answer :

$\rm \cos(1^{ \circ} ) \times \cos(2 ^ {\circ}) \times ... \cos(90^{ \circ} ) = 0$

Learn more :

Trigonometric Table :-

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

$\boxed{\begin{minipage}{6cm} Important Trigonometric identities :- \\ \\ \: \: 1)\:\sin^2\theta+\cos^2\theta=1 \\ \\ 2)\:\sin^2\theta= 1-\cos^2\theta \\ \\ 3)\:\cos^2\theta=1-\sin^2\theta \\ \\ 4)\:1+\cot^2\theta=\text{cosec}^2 \, \theta \\ \\5)\: \text{cosec}^2 \, \theta-\cot^2\theta =1 \\ \\ 6)\:\text{cosec}^2 \, \theta= 1+\cot^2\theta \\\ \\ 7)\:\sec^2\theta=1+\tan^2\theta \\ \\ 8)\:\sec^2\theta-\tan^2\theta=1 \\ \\ 9)\:\tan^2\theta=\sec^2\theta-1 \end{minipage}}$