Question

find the value of 3cot^2 60° +sec^2 45°​

Answer 1

$\;\;\underline{\textbf{\textsf{ Given:-}}}$

• $\sf3\cot^260\degree+\sec^245\degree$

$\;\;\underline{\textbf{\textsf{ To Find :-}}}$

• Value of $\sf3\cot^260\degree+\sec^245\degree$.

$\;\;\underline{\textbf{\textsf{ Solution:-}}}$

$\\ \underline{\textsf{Given that }}$

$\sf3\cot^260\degree+\sec^245\degree$

$\sf\dashrightarrow 3(\cot60\degree)^2+(\sec45\degree)^2$

$\\ \underline{\textsf{ Now, put the values of cot60° and sec45°}}$

$\sf\dashrightarrow 3\times(\dfrac{1}{\sqrt{3}})^2+(\sqrt{2})^2$

$\sf\dashrightarrow 3\times\dfrac{1}{3}+2$

$\sf\dashrightarrow 1+2$

$\sf\dashrightarrow 3$

$\;\;\underline{\textbf{\textsf{ Hence:-}}}$

Value of $\sf3\cot^260\degree+\sec^245\degree$ is 3.

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$\\ \underline{\textsf{Have a look here }}$

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$\begin{array}{ |c |c|c|c|c|c|} \bf\angle A & \bf{0}^{ \circ} & \bf{30}^{ \circ} & \bf{45}^{ \circ} & \bf{60}^{ \circ} & \bf{90}^{ \circ} \\ \\ \rm sin A & 0 & \dfrac{1}{2}& \dfrac{1}{ \sqrt{2} } & \dfrac{ \sqrt{3} }{2} &1 \\ \\ \rm cos \: A & 1 & \dfrac{ \sqrt{3} }{2}& \dfrac{1}{ \sqrt{2} } & \dfrac{1}{2} &0 \\ \\ \rm tan A & 0 & \dfrac{1}{ \sqrt{3} }& 1 & \sqrt{3} & \rm Not \: De fined \\ \\ \rm cosec A & \rm Not \: De fined & 2& \sqrt{2} & \dfrac{2}{ \sqrt{3} } &1 \\ \\ \rm sec A & 1 & \dfrac{2}{ \sqrt{3} }& \sqrt{2} & 2 & \rm Not \: De fined \\ \\ \rm cot A & \rm Not \: De fined & \sqrt{3} & 1 & \dfrac{1}{ \sqrt{3} } & 0 \end{array}$

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Answer 2

Answer:

3 is the correct answer