Question

Prove that $\left( \sec \theta +\cos \theta \right)\left( \sec \theta -\cos \theta \right)={{\tan }^{2}}\theta +{{\sin }^{2}}\theta $.

Answer 1

Hint: Use the fact that $\left( a+b \right)\left( a-b \right)={{a}^{2}}-{{b}^{2}}$. Use ${{\sec }^{2}}\theta =1+{{\tan }^{2}}\theta $ and ${{\sin }^{2}}\theta =1-{{\cos }^{2}}\theta $. Hence simplify LHS and RHS.

Complete step-by-step solution -

Trigonometric ratios:
There are six trigonometric ratios defined on an angle of a right-angled triangle, viz sine, cosine,
tangent, cotangent, secant and cosecant.
The sine of an angle is defined as the ratio of the opposite side to the hypotenuse.
The cosine of an angle is defined as the ratio of the adjacent side to the hypotenuse.
The tangent of an angle is defined as the ratio of the opposite side to the adjacent side.
The cotangent of an angle is defined as the ratio of the adjacent side to the opposite side.
The secant of an angle is defined as the ratio of the hypotenuse to the adjacent side.
The cosecant of an angle is defined as the ratio of the hypotenuse to the adjacent side.
Pythagorean identities:
The following identities are known as Pythagorean identities
${{\sin }^{2}}x+{{\cos }^{2}}x=1,{{\sec }^{2}}x=1+{{\tan }^{2}}x$ and ${{\csc }^{2}}x=1+{{\cot }^{2}}x$ as these are a direct consequence of Pythagoras theorem.
We have LHS $=\left( \sec \theta +\cos \theta \right)\left( \sec \theta -\cos \theta \right)$
We know that $\left( a+b \right)\left( a-b \right)={{a}^{2}}-{{b}^{2}}$.
Using the above formula, we get
LHS $={{\sec }^{2}}\theta -{{\cos }^{2}}\theta $
We know that ${{\sec }^{2}}\theta =1+{{\tan }^{2}}\theta $
Using the above identity, we get
LHS $=1+{{\tan }^{2}}\theta -{{\cos }^{2}}\theta ={{\tan }^{2}}\theta +1-{{\cos }^{2}}\theta $
We know that ${{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1$
Subtracting ${{\cos }^{2}}\theta $ from both sides, we get
${{\sin }^{2}}\theta =1-{{\cos }^{2}}\theta $
Hence we have
LHS $={{\tan }^{2}}\theta +{{\sin }^{2}}\theta $
Hence LHS = RHS.
Q.E.D

Note: Alternatively, we have
RHS $={{\tan }^{2}}\theta +{{\sin }^{2}}\theta $
We know that $\tan \theta =\dfrac{\sin \theta }{\cos \theta }$
Hence we have
$=\dfrac{{{\sin }^{2}}\theta }{{{\cos }^{2}}\theta }+{{\sin }^{2}}\theta $
Taking ${{\sin }^{2}}\theta $ common from both the terms, we get
RHS $={{\sin }^{2}}\theta \left( \dfrac{1}{{{\cos }^{2}}\theta }+1 \right)$
We know that ${{\sin }^{2}}\theta =1-{{\cos }^{2}}\theta $
Hence we get
RHS \[=\dfrac{\left( 1-{{\cos }^{2}}\theta \right)\left( 1+{{\cos }^{2}}\theta \right)}{{{\cos }^{2}}\theta }=\dfrac{1-{{\cos }^{4}}\theta }{{{\cos }^{2}}\theta }\]
Hence we have
RHS $=\dfrac{1}{{{\cos }^{2}}\theta }-\dfrac{{{\cos }^{4}}\theta }{{{\cos }^{2}}\theta }={{\sec }^{2}}\theta -{{\cos }^{2}}\theta =\left( \sec \theta +\cos \theta \right)\left( \sec \theta -\cos \theta \right)$
Hence we get RHS = LHS
Q.E.D