Question

To prove the following identity:
${{(1+\tan \alpha \tan \beta )}^{2}}+{{(\tan \alpha -\tan \beta )}^{2}}={{\sec }^{2}}\alpha {{\sec }^{2}}\beta$

Answer 1

To solve the following trigonometric expression, we should be aware about the basic trigonometric relation between tan$\theta$ and sec$\theta$. This relation is given by- ${{\sec }^{2}}\theta =1+{{\tan }^{2}}\theta$
This property will be put to use after expanding the LHS term.

Complete step-by-step answer:
We have the following expression on the LHS-
=${{(1+\tan \alpha \tan \beta )}^{2}}+{{(\tan \alpha -\tan \beta )}^{2}}$
Now we will use the basic trigonometric identity in the above equation which is given as,
${{(a+b)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab$ and ${{(a-b)}^{2}}={{a}^{2}}+{{b}^{2}}-2ab$
Using the above identities we get,
=$1+{{(\tan \alpha \tan \beta )}^{2}}+2\tan \alpha \tan \beta +{{\tan }^{2}}\alpha +{{\tan }^{2}}\beta -2\tan \alpha \tan \beta$
Cancelling the terms we get,
=$1+{{(\tan \alpha \tan \beta )}^{2}}+{{\tan }^{2}}\alpha +{{\tan }^{2}}\beta$
=$1+{{\tan }^{2}}\alpha {{\tan }^{2}}\beta +{{\tan }^{2}}\alpha +{{\tan }^{2}}\beta$
Taking ${{\tan }^{2}}\beta$ common we get,
=$1+{{\tan }^{2}}\alpha +{{\tan }^{2}}\beta (1+{{\tan }^{2}}\alpha )$
Making necessary arrangements we have,
=$(1+{{\tan }^{2}}\beta )(1+{{\tan }^{2}}\alpha )$
Now, we use the property that ${{\sec }^{2}}\theta =1+{{\tan }^{2}}\theta$. Thus, we have,
=$({{\sec }^{2}}\beta )({{\sec }^{2}}\alpha )$
=${{\sec }^{2}}\alpha {{\sec }^{2}}\beta$
We can clearly see that this is the same as the RHS term. Thus, LHS=RHS, hence, proved.

Note: While solving problems related to trigonometric expression, there are several manipulations and re-grouping required while arriving at the final answer. As a prerequisite, one should be well aware of the basic trigonometric relations and properties such as the one mentioned in the above problem. However, in addition to the usage of these properties and relations, one needs to do several manipulations and use of other subjects to solve the trigonometric expressions. As an alternative, one can also start from the RHS of the problem by expanding ${{\sec }^{2}}\theta =1+{{\tan }^{2}}\theta$ in the RHS and solving to arrive at the LHS. Another alternative to solve this is by expressing tanx and secx in terms of sinx and cosx. Thus, on the LHS, we get,
=${{\left( 1+\dfrac{\sin \alpha \sin \beta }{\cos \alpha \cos \beta } \right)}^{2}}+{{\left( \dfrac{\sin \alpha }{\cos \alpha }-\dfrac{\sin \beta }{\cos \beta } \right)}^{2}}$
= ${{\left( \dfrac{\cos \alpha \cos \beta +\sin \alpha \sin \beta }{\cos \alpha \cos \beta } \right)}^{2}}+{{\left( \dfrac{\sin \alpha \cos \beta -\cos \alpha \sin \beta }{\cos \alpha \cos \beta } \right)}^{2}}$
Now, we know that, $\cos \alpha \cos \beta +\sin \alpha \sin \beta =\cos (\alpha -\beta )$ and $\sin \alpha \cos \beta - \cos \alpha \sin \beta =\sin (\alpha -\beta )$. We substitute this in the above expression of LHS, to simplify further,
= ${{\left( \dfrac{\cos (\alpha - \beta )}{\cos \alpha \cos \beta } \right)}^{2}}+{{\left( \dfrac{\sin (\alpha - \beta )}{\cos \alpha \cos \beta } \right)}^{2}}$
=$\dfrac{{{\cos }^{2}}(\alpha -\beta )+{{\sin }^{2}}(\alpha -\beta )}{\cos \alpha \cos \beta }$
Now, using the property that ${{\cos }^{2}}x+{{\sin }^{2}}x=1$, we get,
=$\dfrac{1}{{{\cos}^{2}} \alpha {{\cos}^{2}} \beta }$
Now, again converting $\sec x$ into $\cos x$, we get
= ${{\sec}^{2}} \alpha {{\sec}^{2}} \beta$
Which is the same as RHS, thus we could prove that LHS=RHS using the above technique too.