Question

Prove the following expression:

\[{{\left( \sec A\sec B+\tan A\tan B \right)}^{2}}-{{\left( \sec A\tan B+\tan A\sec B \right)}^{2}}=1\]

Answer 1

Hint: To prove this expression, if we start from left-hand side, then we need to know a few trigonometric identities like \[\sec \theta =\dfrac{1}{\cos \theta }\], \[\tan \theta =\dfrac{\sin \theta }{\cos \theta }\]. After converting sec and tan in terms of sin and cos then we have to simply apply algebraic properties to simplify it.

Complete step-by-step answer:

In this question, we are asked to prove that

\[{{\left( \sec A\sec B+\tan A\tan B \right)}^{2}}-{{\left( \sec A\tan B+\tan A\sec B \right)}^{2}}=1\]

To prove this expression, we will first consider the left-hand side of the expression. So, we can write it as,

\[LHS={{\left( \sec A\sec B+\tan A\tan B \right)}^{2}}-{{\left( \sec A\tan B+\tan A\sec B \right)}^{2}}\]

Now, we know that \[\sec \theta =\dfrac{1}{\cos \theta }\] and \[\tan \theta =\dfrac{\sin \theta }{\cos \theta }\]. So, by using these formulas, we will get LHS as,

\[LHS={{\left( \dfrac{1}{\cos A\cos B}+\dfrac{\sin A\sin B}{\cos A\cos B} \right)}^{2}}-{{\left( \dfrac{\sin B}{\cos A\cos B}+\dfrac{\sin A}{\cos A\cos B} \right)}^{2}}\]

Now, we will take the LCM of each term separately. So, we will get,

\[LHS={{\left( \dfrac{1+\sin A\sin B}{\cos A\cos B} \right)}^{2}}-{{\left( \dfrac{\sin B+\sin A}{\cos A\cos B} \right)}^{2}}\]

Now, we know that when we have to simplify a fractional term, we take its LCM. So, we get,
\[LHS=\dfrac{{{\left( 1+\sin A\sin B \right)}^{2}}-{{\left( \sin B+\sin A \right)}^{2}}}{{{\left( \cos A\cos B \right)}^{2}}}\]

Now, we know that \[{{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab\]. So, we will get LHS as,

\[LHS=\dfrac{\left( 1+{{\sin }^{2}}A{{\sin }^{2}}B+2\sin A\sin B \right)-\left( {{\sin }^{2}}A+{{\sin }^{2}}B+2\sin A\sin B \right)}{{{\cos }^{2}}A{{\cos }^{2}}B}\]

\[LHS=\dfrac{\left( 1+{{\sin }^{2}}A{{\sin }^{2}}B+2\sin A\sin B-{{\sin }^{2}}A-{{\sin }^{2}}B-2\sin A\sin B \right)}{{{\cos }^{2}}A{{\cos }^{2}}B}\]

Now, we know that the equal terms with opposite signs get canceled out. So, we get, LHS as
\[LHS=\dfrac{1-{{\sin }^{2}}A-{{\sin }^{2}}B+{{\sin }^{2}}A{{\sin }^{2}}B}{{{\cos }^{2}}A{{\cos }^{2}}B}\]

Now, we know that \[\left( -{{\sin }^{2}}B+{{\sin }^{2}}A{{\sin }^{2}}B \right)\] can be expressed as \[-{{\sin }^{2}}B\left( 1-{{\sin }^{2}}A \right)\]. Therefore, we can write LHS as
\[LHS=\dfrac{\left( 1-{{\sin }^{2}}A \right)-{{\sin }^{2}}B\left( 1-{{\sin }^{2}}A \right)}{{{\cos }^{2}}A{{\cos }^{2}}B}\]

Now, we will take \[1-{{\sin }^{2}}A\] common from the numerator, so we get

\[LHS=\dfrac{\left( 1-{{\sin }^{2}}A \right)\left( 1-{{\sin }^{2}}B \right)}{{{\cos }^{2}}A{{\cos }^{2}}B}\]

Now, we know that \[{{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1\] which means \[1-{{\sin }^{2}}\theta ={{\cos }^{2}}\theta \]. Therefore, we get LHS as

\[LHS=\dfrac{{{\cos }^{2}}A{{\cos }^{2}}B}{{{\cos }^{2}}A{{\cos }^{2}}B}\]

And we know that the common terms in both the numerator and denominator get canceled out. Therefore, we get,

LHS = 1

LHS = RHS

Hence proved


Note: The possible mistakes one can do while solving the question is by opening up the brackets as the very first step. That is not wrong, but that will increase the complexity of the solution which will increase the chances of calculation mistakes.