
Prove that $si{{n}^{2}}\left( A \right)tan\left( A \right)+co{{s}^{2}}\left( A \right)cot\left( A \right)+2sin\left( A \right)cos\left( A \right)=tan\left( A \right)+cot\left( A \right)$
Answer
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Hint: We are given two expressions involving trigonometric terms and we need to prove that one is equal to the other. For this we use several trigonometric formulae and we try to simplify the equation as much as possible. We will start with the left side and we then perform some operations on it to turn it into the expression written on the right hand side.
Complete step-by-step answer:
We start with the left hand side, we have
$LHS=si{{n}^{2}}(A)tan(A)+co{{s}^{2}}(A)cot(A)+2Sin(A)Cos(A)$
We see the terms involving squares of sine and cosine, so we use the following formula here:
$si{{n}^{2}}(A)+co{{s}^{2}}(A)=1$
And this holds as true for any$A$.
So, we put $si{{n}^{2}}(A)=1-co{{s}^{2}}(A)$and $co{{s}^{2}}(A)=1-si{{n}^{2}}(A)$in the above equation. So, we get:
$\left( 1-co{{s}^{2}}A \right)tan(A)+\left( 1-si{{n}^{2}}A \right)cot(A)+2sin(A)cos(A)$
Now, we resolve the brackets and we get the following:
$tan(A)-co{{s}^{2}}(A)tan(A)+cot(A)-si{{n}^{2}}(A)cot(A)+2sin(A)cos(A)$
Now, we have the square of cosine multiplied by tan, so we can use the following formula:
$tan(A)=\frac{Sin(A)}{Cos(A)}$
Similarly we have the term involving sin squared multiplied by cot, so we use the following too:
$Cot(A)=\frac{Cos(A)}{Sin(A)}$
Plugging these in the expression, we get the following:
$= tan(A)-\left( co{{s}^{2}}(A)\times \frac{sin(A)}{cos(A)} \right)+cot(A)-\left( si{{n}^{2}}(A)\times \frac{cos(A)}{sin(A)} \right)+2sin(A)cos(A)$
$= tan(A)-sin(A)cos(A)+cot(A)-sin(A)cos(A)+2sin(A)cos(A)$
Now, we keep the terms involving tan and cot aside and the terms involving sin and cosine aside, we get the following:
$= tan(A)+cot(A)-2sin(A)cos(A)+2sin(A)cos(A)$
$2sin(A)cos(A)$gets cancelled out from both sides and we finally get:
$tan(A)+cot(A)=RHS$
Hence, proved.
Note: Most of the times, while using the trigonometric formulae, we may plug in the wrong equations. So, the trigonometric formulae must be learned properly so that you are able to prove the quality of expressions without any mistakes. Moreover, always start with the side that is complex. As in this question, if you start with the left hand side then you will have no operation to perform and you will get stuck, so always start with the complex side and try to reduce it to the expression that doesn’t involve much terms.
Complete step-by-step answer:
We start with the left hand side, we have
$LHS=si{{n}^{2}}(A)tan(A)+co{{s}^{2}}(A)cot(A)+2Sin(A)Cos(A)$
We see the terms involving squares of sine and cosine, so we use the following formula here:
$si{{n}^{2}}(A)+co{{s}^{2}}(A)=1$
And this holds as true for any$A$.
So, we put $si{{n}^{2}}(A)=1-co{{s}^{2}}(A)$and $co{{s}^{2}}(A)=1-si{{n}^{2}}(A)$in the above equation. So, we get:
$\left( 1-co{{s}^{2}}A \right)tan(A)+\left( 1-si{{n}^{2}}A \right)cot(A)+2sin(A)cos(A)$
Now, we resolve the brackets and we get the following:
$tan(A)-co{{s}^{2}}(A)tan(A)+cot(A)-si{{n}^{2}}(A)cot(A)+2sin(A)cos(A)$
Now, we have the square of cosine multiplied by tan, so we can use the following formula:
$tan(A)=\frac{Sin(A)}{Cos(A)}$
Similarly we have the term involving sin squared multiplied by cot, so we use the following too:
$Cot(A)=\frac{Cos(A)}{Sin(A)}$
Plugging these in the expression, we get the following:
$= tan(A)-\left( co{{s}^{2}}(A)\times \frac{sin(A)}{cos(A)} \right)+cot(A)-\left( si{{n}^{2}}(A)\times \frac{cos(A)}{sin(A)} \right)+2sin(A)cos(A)$
$= tan(A)-sin(A)cos(A)+cot(A)-sin(A)cos(A)+2sin(A)cos(A)$
Now, we keep the terms involving tan and cot aside and the terms involving sin and cosine aside, we get the following:
$= tan(A)+cot(A)-2sin(A)cos(A)+2sin(A)cos(A)$
$2sin(A)cos(A)$gets cancelled out from both sides and we finally get:
$tan(A)+cot(A)=RHS$
Hence, proved.
Note: Most of the times, while using the trigonometric formulae, we may plug in the wrong equations. So, the trigonometric formulae must be learned properly so that you are able to prove the quality of expressions without any mistakes. Moreover, always start with the side that is complex. As in this question, if you start with the left hand side then you will have no operation to perform and you will get stuck, so always start with the complex side and try to reduce it to the expression that doesn’t involve much terms.
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