Math, asked by PRINCESS100001, 2 months ago

Q1.Heya everyone.
Find the value of the following..
\sf \cos(1^{ \circ} ) \times \cos(2 ^ {\circ}) \times ... \cos(90^{ \circ}

Answers

Answered by PRINCE100001
11

Step-by-step explanation:

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</p><p>

Answer :

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

Learn more :

Trigonometric Table :-

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

\begin{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}}\end{gathered}

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