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Question: Prove the following statement. \[\left( \dfrac{1}{{{\sec }^{2}}\alpha -{{\cos }^{2}}\alpha }+\dfra...

Prove the following statement.
(1sec⁡2α−cos⁡2α+1cosec⁡2α−sin⁡2α)sin⁡2αcos⁡2α=1−cos⁡2αsin⁡2α2+cos⁡2αsin⁡2α\left( \dfrac{1}{{{\sec }^{2}}\alpha -{{\cos }^{2}}\alpha }+\dfrac{1}{{{\operatorname{cosec}}^{2}}\alpha -{{\sin }^{2}}\alpha } \right){{\sin }^{2}}\alpha {{\cos }^{2}}\alpha =\dfrac{1-{{\cos }^{2}}\alpha {{\sin }^{2}}\alpha }{2+{{\cos }^{2}}\alpha {{\sin }^{2}}\alpha }

Explanation

Solution

Hint: To prove this question, we should know that sec α can be written as 1cos⁡α\dfrac{1}{\cos \alpha } and cosec α can be written as 1sin⁡α\dfrac{1}{\sin \alpha }. Also, we should know a few algebraic identities like, a2−b2=(a−b)(a+b){{a}^{2}}-{{b}^{2}}=\left( a-b \right)\left( a+b \right) and (a+b)2=a2+b2+2ab{{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab. By using these identities we can prove the statement.
Complete step-by-step answer:
In this question, we have been asked to prove (1sec⁡2α−cos⁡2α+1cosec⁡2α−sin⁡2α)sin⁡2αcos⁡2α=1−cos⁡2αsin⁡2α2+cos⁡2αsin⁡2α\left( \dfrac{1}{{{\sec }^{2}}\alpha -{{\cos }^{2}}\alpha }+\dfrac{1}{{{\operatorname{cosec}}^{2}}\alpha -{{\sin }^{2}}\alpha } \right){{\sin }^{2}}\alpha {{\cos }^{2}}\alpha =\dfrac{1-{{\cos }^{2}}\alpha {{\sin }^{2}}\alpha }{2+{{\cos }^{2}}\alpha {{\sin }^{2}}\alpha }. To prove this relation, we will first consider the left hand side of the relation. So, we can write it as,
LHS=(1sec⁡2α−cos⁡2α+1cosec⁡2α−sin⁡2α)sin⁡2αcos⁡2αLHS=\left( \dfrac{1}{{{\sec }^{2}}\alpha -{{\cos }^{2}}\alpha }+\dfrac{1}{{{\operatorname{cosec}}^{2}}\alpha -{{\sin }^{2}}\alpha } \right){{\sin }^{2}}\alpha {{\cos }^{2}}\alpha
Now, we know that sec⁡α=1cos⁡α\sec \alpha =\dfrac{1}{\cos \alpha } and cosec⁡α=1sin⁡α\operatorname{cosec}\alpha =\dfrac{1}{\sin \alpha }. So, we will get the LHS as,
LHS=(11cos⁡2α−cos⁡2α+11sin⁡2α−sin⁡2α)sin⁡2αcos⁡2αLHS=\left( \dfrac{1}{\dfrac{1}{{{\cos }^{2}}\alpha }-{{\cos }^{2}}\alpha }+\dfrac{1}{\dfrac{1}{{{\sin }^{2}}\alpha }-{{\sin }^{2}}\alpha } \right){{\sin }^{2}}\alpha {{\cos }^{2}}\alpha
Now, we will take the LCM of each term of the LHS. So, we will get,
LHS=(11−cos⁡4αcos⁡2α+11−sin⁡4αsin⁡2α)sin⁡2αcos⁡2α LHS=(cos⁡2α1−cos⁡4α+sin⁡2α1−sin⁡4α)sin⁡2αcos⁡2 \begin{aligned} & LHS=\left( \dfrac{1}{\dfrac{1-{{\cos }^{4}}\alpha }{{{\cos }^{2}}\alpha }}+\dfrac{1}{\dfrac{1-{{\sin }^{4}}\alpha }{{{\sin }^{2}}\alpha }} \right){{\sin }^{2}}\alpha {{\cos }^{2}}\alpha \\\ & LHS=\left( \dfrac{{{\cos }^{2}}\alpha }{1-{{\cos }^{4}}\alpha }+\dfrac{{{\sin }^{2}}\alpha }{1-{{\sin }^{4}}\alpha } \right){{\sin }^{2}}\alpha {{\cos }^{2}} \\\ \end{aligned}
We can write cos⁡4α{{\cos }^{4}}\alpha as (cos⁡2α)2{{\left( {{\cos }^{2}}\alpha \right)}^{2}} and sin⁡4α{{\sin }^{4}}\alpha as (sin⁡2α)2{{\left( {{\sin }^{2}}\alpha \right)}^{2}}. So, we will get, LHS=(cos⁡2α1−(cos⁡2α)2+sin⁡2α1−(sin⁡2α)2)sin⁡2αcos⁡2LHS=\left( \dfrac{{{\cos }^{2}}\alpha }{1-{{\left( {{\cos }^{2}}\alpha \right)}^{2}}}+\dfrac{{{\sin }^{2}}\alpha }{1-{{\left( {{\sin }^{2}}\alpha \right)}^{2}}} \right){{\sin }^{2}}\alpha {{\cos }^{2}}
Now, we know that a2−b2=(a−b)(a+b){{a}^{2}}-{{b}^{2}}=\left( a-b \right)\left( a+b \right). So, on applying this in the above expression we will get,
LHS=(cos⁡2α(1−cos⁡2α)(1+cos⁡2α)+sin⁡2α(1−sin⁡2α)(1+sin⁡2α))sin⁡2αcos⁡2LHS=\left( \dfrac{{{\cos }^{2}}\alpha }{\left( 1-{{\cos }^{2}}\alpha \right)\left( 1+{{\cos }^{2}}\alpha \right)}+\dfrac{{{\sin }^{2}}\alpha }{\left( 1-{{\sin }^{2}}\alpha \right)\left( 1+{{\sin }^{2}}\alpha \right)} \right){{\sin }^{2}}\alpha {{\cos }^{2}}
Now, we also know that sin⁡2α+cos⁡2α=1{{\sin }^{2}}\alpha +{{\cos }^{2}}\alpha =1. So, we can write 1−cos⁡2α=sin⁡2α1-{{\cos }^{2}}\alpha ={{\sin }^{2}}\alpha and 1−sin⁡2α=cos⁡2α1-{{\sin }^{2}}\alpha ={{\cos }^{2}}\alpha . Therefore, we will get the LHS as,
LHS=(cos⁡2αsin⁡2α(1+cos⁡2α)+sin⁡2αcos⁡2α(1+sin⁡2α))sin⁡2αcos⁡2LHS=\left( \dfrac{{{\cos }^{2}}\alpha }{{{\sin }^{2}}\alpha \left( 1+{{\cos }^{2}}\alpha \right)}+\dfrac{{{\sin }^{2}}\alpha }{{{\cos }^{2}}\alpha \left( 1+{{\sin }^{2}}\alpha \right)} \right){{\sin }^{2}}\alpha {{\cos }^{2}}
Now, we will open the brackets to simply it,
LHS=(cos⁡4αsin⁡2αsin⁡2α(1+cos⁡2α)+sin⁡4αcos⁡2αcos⁡2α(1+sin⁡2α))LHS=\left( \dfrac{{{\cos }^{4}}\alpha {{\sin }^{2}}\alpha }{{{\sin }^{2}}\alpha \left( 1+{{\cos }^{2}}\alpha \right)}+\dfrac{{{\sin }^{4}}\alpha {{\cos }^{2}}\alpha }{{{\cos }^{2}}\alpha \left( 1+{{\sin }^{2}}\alpha \right)} \right)
We know that common terms get cancelled out, so we get,
LHS=cos⁡4α(1+cos⁡2α)+sin⁡4α(1+sin⁡2α)LHS=\dfrac{{{\cos }^{4}}\alpha }{\left( 1+{{\cos }^{2}}\alpha \right)}+\dfrac{{{\sin }^{4}}\alpha }{\left( 1+{{\sin }^{2}}\alpha \right)}
Now, we will take the LCM of both the terms. So, we will get,

& LHS=\dfrac{{{\cos }^{4}}\alpha \left( 1+{{\sin }^{2}}\alpha \right)+{{\sin }^{4}}\alpha \left( 1+{{\cos }^{2}}\alpha \right)}{\left( 1+{{\cos }^{2}}\alpha \right)\left( 1+{{\sin }^{2}}\alpha \right)} \\\ & LHS=\dfrac{{{\cos }^{4}}\alpha +{{\cos }^{4}}\alpha {{\sin }^{2}}\alpha +{{\sin }^{4}}\alpha +{{\sin }^{4}}\alpha {{\cos }^{2}}\alpha }{1+{{\cos }^{2}}\alpha +{{\sin }^{2}}\alpha +{{\cos }^{2}}\alpha {{\sin }^{2}}\alpha } \\\ \end{aligned}$$ Now, we can see that $${{\cos }^{4}}\alpha {{\sin }^{2}}\alpha +{{\sin }^{4}}\alpha {{\cos }^{2}}\alpha $$ can be written as $${{\sin }^{2}}\alpha {{\cos }^{2}}\alpha \left( {{\cos }^{2}}\alpha +{{\sin }^{2}}\alpha \right)$$. So, we will get, $$LHS=\dfrac{{{\cos }^{4}}\alpha +{{\sin }^{4}}\alpha +{{\sin }^{2}}\alpha {{\cos }^{2}}\alpha \left( {{\cos }^{2}}\alpha +{{\sin }^{2}}\alpha \right)}{1+{{\cos }^{2}}\alpha +{{\sin }^{2}}\alpha +{{\cos }^{2}}\alpha {{\sin }^{2}}\alpha }$$ We also know that ${{\sin }^{2}}\alpha +{{\cos }^{2}}\alpha =1$ so, we can write the LHS as, $$\begin{aligned} & LHS=\dfrac{{{\cos }^{4}}\alpha +{{\sin }^{4}}\alpha +{{\sin }^{2}}\alpha {{\cos }^{2}}\alpha \left( 1 \right)}{1+1+{{\sin }^{2}}\alpha {{\cos }^{2}}\alpha } \\\ & LHS=\dfrac{{{\cos }^{4}}\alpha +{{\sin }^{4}}\alpha +{{\sin }^{2}}\alpha {{\cos }^{2}}\alpha }{2+{{\sin }^{2}}\alpha {{\cos }^{2}}\alpha } \\\ \end{aligned}$$ Now, we know that ${{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab$. So, we can say that ${{a}^{2}}+{{b}^{2}}={{\left( a+b \right)}^{2}}-2ab$. Therefore, we can write $${{\cos }^{4}}\alpha +{{\sin }^{4}}\alpha $$ as $${{\left( {{\cos }^{2}}\alpha +{{\sin }^{2}}\alpha \right)}^{2}}-2{{\cos }^{2}}\alpha {{\sin }^{2}}\alpha $$. So, we will get, $$\begin{aligned} & LHS=\dfrac{{{\left( {{\cos }^{2}}\alpha +{{\sin }^{2}}\alpha \right)}^{2}}-2{{\sin }^{2}}\alpha {{\cos }^{2}}\alpha +{{\sin }^{2}}\alpha {{\cos }^{2}}\alpha }{2+{{\sin }^{2}}\alpha {{\cos }^{2}}\alpha } \\\ & LHS=\dfrac{{{\left( {{\cos }^{2}}\alpha +{{\sin }^{2}}\alpha \right)}^{2}}-{{\sin }^{2}}\alpha {{\cos }^{2}}\alpha }{2+{{\sin }^{2}}\alpha {{\cos }^{2}}\alpha } \\\ \end{aligned}$$ Now, we will again put ${{\sin }^{2}}\alpha +{{\cos }^{2}}\alpha =1$. So, we will get, $$\begin{aligned} & LHS=\dfrac{1-{{\sin }^{2}}\alpha {{\cos }^{2}}\alpha }{2+{{\sin }^{2}}\alpha {{\cos }^{2}}\alpha } \\\ & LHS=RHS \\\ \end{aligned}$$ Hence proved. Note: In this question, there are high possibilities that a student may make calculation mistakes. Also, they could make a mistake while applying the algebraic and trigonometric formulas. So, the students have to be careful while solving the question.