Question

Prove that \[\sec \left( {270^\circ - A} \right)\sec \left( {90^\circ - A} \right) - \tan \left( {270^\circ - A} \right)\tan \left( {90^\circ + A} \right) + 1 = 0\].

Answer 1

Hint:
Here, we will use different trigonometric identities and formulas to prove the given equation. We will convert secant to cosecant and tangent to cotangent. Then, we will apply the trigonometric identity i.e. multiple of \[90^\circ \] and simplify the left hand side of the given equation.
Formula Used: We will use the formulas \[{\rm{cose}}{{\rm{c}}^2}\theta - {\cot ^2}\theta = 1\], \[\sec \left( {90^\circ - \theta } \right) = {\rm{cosec }}\theta \] and\[\tan \left( {90^\circ - \theta } \right) = \cot \theta \] to solve the question.

Complete step by step solution:
First, we will convert the secants to cosecants.
We will rewrite the expression \[\sec \left( {270^\circ - A} \right)\sec \left( {90^\circ - A} \right)\].
Rewriting the angle \[270^\circ \] as \[180^\circ + 90^\circ \], we get
\[\begin{array}{c} \Rightarrow \sec \left( {270^\circ - A} \right)\sec \left( {90^\circ - A} \right) = \sec \left( {180^\circ + 90^\circ - A} \right)\sec \left( {90^\circ - A} \right)\\ \Rightarrow \sec \left( {270^\circ - A} \right)\sec \left( {90^\circ - A} \right) = \sec \left( {180^\circ + \left( {90^\circ - A} \right)} \right)\sec \left( {90^\circ - A} \right)\end{array}\]
We know that secant is negative in the third quadrant. Therefore, \[\sec \left( {180^\circ + \theta } \right) = - \sec \theta \] (secant does not become cosecant because \[180^\circ \] is an even multiple of \[90^\circ \]).
Therefore, we get
\[\begin{array}{l} \Rightarrow \sec \left( {270^\circ - A} \right)\sec \left( {90^\circ - A} \right) = - \sec \left( {90^\circ - A} \right)\sec \left( {90^\circ - A} \right)\\ \Rightarrow \sec \left( {270^\circ - A} \right)\sec \left( {90^\circ - A} \right) = - {\sec ^2}\left( {90^\circ - A} \right)\end{array}\]
We know that the secant and cosecant of two complementary angles follow the relation \[\sec \left( {90^\circ - \theta } \right) = {\rm{cosec }}\theta \].
Substituting \[\sec \left( {90^\circ - A} \right) = {\rm{cosec }}A\] in the expression \[ - {\sec ^2}\left( {90^\circ - A} \right)\], we get
\[ \Rightarrow \sec \left( {270^\circ - A} \right)\sec \left( {90^\circ - A} \right) = - {\rm{cose}}{{\rm{c}}^2}A\]
Now, we will convert the tangents to cotangents.
We will rewrite the expression \[\tan \left( {270^\circ - A} \right)\tan \left( {90^\circ + A} \right)\].
Rewriting the angle \[270^\circ \] as \[180^\circ + 90^\circ \], we get
\[\begin{array}{c} \Rightarrow \tan \left( {270^\circ - A} \right)\tan \left( {90^\circ + A} \right) = \tan \left( {180^\circ + 90^\circ - A} \right)\tan \left( {90^\circ + A} \right)\\ \Rightarrow \tan \left( {270^\circ - A} \right)\tan \left( {90^\circ + A} \right) = \tan \left( {180^\circ + \left( {90^\circ - A} \right)} \right)\tan \left( {90^\circ + A} \right)\end{array}\]
We know that tangent is positive in the third quadrant. Therefore, \[\tan \left( {180^\circ + \theta } \right) = \tan \theta \] (tangent does not become cotangent because \[180^\circ \] is an even multiple of \[90^\circ \]).
Therefore, we get
\[ \Rightarrow \tan \left( {270^\circ - A} \right)\tan \left( {90^\circ + A} \right) = \tan \left( {90^\circ - A} \right)\tan \left( {90^\circ + A} \right)\]
We know that the tangent and cotangent of two complementary angles follow the relation \[\tan \left( {90^\circ - \theta } \right) = \cot \theta \].
Substituting \[\tan \left( {90^\circ - A} \right) = \cot A\] in the expression \[\tan \left( {90^\circ - A} \right)\tan \left( {90^\circ + A} \right)\], we get
\[ \Rightarrow \tan \left( {270^\circ - A} \right)\tan \left( {90^\circ + A} \right) = \cot A\tan \left( {90^\circ + A} \right)\]
The tangent of an angle is negative in the second quadrant. Therefore, \[\tan \left( {90^\circ + A} \right) = - \cot A\] (tangent becomes cotangent because \[90^\circ \] is an odd multiple of \[90^\circ \]).
Substituting \[\tan \left( {90^\circ + A} \right) = - \cot A\], we get
\[\begin{array}{l} \Rightarrow \tan \left( {270^\circ - A} \right)\tan \left( {90^\circ + A} \right) = \cot A\left( { - \cot A} \right)\\ \Rightarrow \tan \left( {270^\circ - A} \right)\tan \left( {90^\circ + A} \right) = - {\cot ^2}A\end{array}\]
Now, we will solve the left hand side of the equation required to be proved.
Substitute \[\sec \left( {270^\circ - A} \right)\sec \left( {90^\circ - A} \right) = - {\rm{cose}}{{\rm{c}}^2}A\] and \[\tan \left( {270^\circ - A} \right)\tan \left( {90^\circ + A} \right) = - {\cot ^2}A\] in the left hand side of the equation, we get
\[\begin{array}{c}\sec \left( {270^\circ - A} \right)\sec \left( {90^\circ - A} \right) - \tan \left( {270^\circ - A} \right)\tan \left( {90^\circ + A} \right) + 1 = - {\rm{cose}}{{\rm{c}}^2}A - \left( { - {{\cot }^2}A} \right) + 1\\ = - {\rm{cose}}{{\rm{c}}^2}A + {\cot ^2}A + 1\\ = - \left( {{\rm{cose}}{{\rm{c}}^2}A - {{\cot }^2}A} \right) + 1\end{array}\]
Using the trigonometric identity \[{\rm{cose}}{{\rm{c}}^2}\theta - {\cot ^2}\theta = 1\], we get
\[\begin{array}{l} \Rightarrow \sec \left( {270^\circ - A} \right)\sec \left( {90^\circ - A} \right) - \tan \left( {270^\circ - A} \right)\tan \left( {90^\circ + A} \right) + 1 = - 1 + 1\\ \Rightarrow \sec \left( {270^\circ - A} \right)\sec \left( {90^\circ - A} \right) - \tan \left( {270^\circ - A} \right)\tan \left( {90^\circ + A} \right) + 1 = 0\end{array}\]
\[\therefore\] The equation has been proved.

Note:
It is important for us to remember how to convert the secants to cosecant and the tangents to cotangents. For example, \[\tan \left( {180^\circ + \theta } \right) = \tan \theta \] because \[180^\circ \] is an even multiple of \[90^\circ \]. If it was \[\tan \left( {270^\circ + \theta } \right)\], it would become cotangent. We can also solve this question by converting the given expression to an expression of sine and cosine. Then, apply the formula for sine and cosine of sum/difference of two angles to simplify the expression. But the process will become more complicated to solve.