Answer
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Hint: To solve this question we use basic trigonometric identities and formulas and use rearrangements of the obtained terms. Basic trigonometric identities used are as, \[{{\sin }^{2}}A+{{\cos }^{2}}A=1\] and \[2(\cos A)(sinA)=\sin 2A\]
Complete step-by-step answer:
Given equation is tanA + cotA = 2cosec2A. We have to prove that the equation is true or that it holds.
Consider tanA + cotA = 2cosec2A………(i)
Taking LHS of the equation (i) gives,
tanA + cotA
We know that \[\operatorname{tanA}=\left\{ \dfrac{\sin A}{\cos A} \right\}\], and \[cotA=\left\{ \dfrac{\cos A}{\sin A} \right\}\],
Substituting these formulae in the above expression,
\[\Rightarrow \]\[\tan A+cotA=\dfrac{\sin A}{\cos A}+\dfrac{\cos A}{\sin A}\]
Taking LCM of the right-hand side,
\[\Rightarrow \tan A+\cot A=\dfrac{{{\sin }^{2}}A+{{\cos }^{2}}A}{(\cos A)(sinA)}\]
Also, we know that, \[{{\sin }^{2}}A+{{\cos }^{2}}A=1\].
Substituting this value in the above equation,
\[\Rightarrow \tan A+\cot A=\dfrac{1}{(\cos A)(sinA)}\]
When we multiply and divide the right-hand side value by 2, we get,
\[\Rightarrow \tan A+\cot A=\dfrac{2}{2(\cos A)(sinA)}\]
Also, we know that \[2(\cos A)(sinA)=\sin 2A\].
Substituting this value in the above obtained equation,
\[\begin{align}
& \Rightarrow \tan A+\cot A=\dfrac{2}{\sin 2A} \\
& \Rightarrow \tan A+\cot A=2cosec2A \\
\end{align}\]
Therefore, we obtained that the left-hand side becomes equal to the right-hand side.
Hence the statement in the question is proved.
Note: While solving these types of equations where we have to equate the left hand side and the right hand side, always remember that while taking the left hand side in the starting of the question do not tamper with it in the middle of the question, this could be a possibility of error. Just only go on solving with the right-hand side to obtain the equality between the left-hand side and right-hand side.
Complete step-by-step answer:
Given equation is tanA + cotA = 2cosec2A. We have to prove that the equation is true or that it holds.
Consider tanA + cotA = 2cosec2A………(i)
Taking LHS of the equation (i) gives,
tanA + cotA
We know that \[\operatorname{tanA}=\left\{ \dfrac{\sin A}{\cos A} \right\}\], and \[cotA=\left\{ \dfrac{\cos A}{\sin A} \right\}\],
Substituting these formulae in the above expression,
\[\Rightarrow \]\[\tan A+cotA=\dfrac{\sin A}{\cos A}+\dfrac{\cos A}{\sin A}\]
Taking LCM of the right-hand side,
\[\Rightarrow \tan A+\cot A=\dfrac{{{\sin }^{2}}A+{{\cos }^{2}}A}{(\cos A)(sinA)}\]
Also, we know that, \[{{\sin }^{2}}A+{{\cos }^{2}}A=1\].
Substituting this value in the above equation,
\[\Rightarrow \tan A+\cot A=\dfrac{1}{(\cos A)(sinA)}\]
When we multiply and divide the right-hand side value by 2, we get,
\[\Rightarrow \tan A+\cot A=\dfrac{2}{2(\cos A)(sinA)}\]
Also, we know that \[2(\cos A)(sinA)=\sin 2A\].
Substituting this value in the above obtained equation,
\[\begin{align}
& \Rightarrow \tan A+\cot A=\dfrac{2}{\sin 2A} \\
& \Rightarrow \tan A+\cot A=2cosec2A \\
\end{align}\]
Therefore, we obtained that the left-hand side becomes equal to the right-hand side.
Hence the statement in the question is proved.
Note: While solving these types of equations where we have to equate the left hand side and the right hand side, always remember that while taking the left hand side in the starting of the question do not tamper with it in the middle of the question, this could be a possibility of error. Just only go on solving with the right-hand side to obtain the equality between the left-hand side and right-hand side.
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