Question

Prove that ${{\tan }^{2}}A{{\sec }^{2}}B-{{\sec }^{2}}A{{\tan }^{2}}B={{\tan }^{2}}A-{{\tan }^{2}}B$.

Answer 1

Hint: Use the trigonometric identity ${{\sec }^{2}}x=1+{{\tan }^{2}}x$. Hence, express all sec terms in terms of tan and simplify. Alternatively, convert into sines and cosines and simplify.

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.
Observe that sine and cosecant are multiplicative inverses of each other, cosine and secant are multiplicative inverses of each other, and tangent and cotangent are multiplicative inverses of each other.
Hence $\sec x\cos x=1,\sin x\csc x=1$ and $\tan x\cot x=1$.
Trigonometric identities:
Pythagorean Identities:
${{\sin }^{2}}x+{{\cos }^{2}}x=1,{{\sec }^{2}}x=1+{{\tan }^{2}}x$ and ${{\csc }^{2}}x=1+{{\cot }^{2}}x$. These identities are a consequence of Pythagoras theorem in a right-angled triangle.
Now, we have
LHS $={{\tan }^{2}}A{{\sec }^{2}}B-{{\tan }^{2}}B{{\sec }^{2}}A$
Using ${{\sec }^{2}}x=1+{{\tan }^{2}}x$, we get
LHS $={{\tan }^{2}}A\left( 1+{{\tan }^{2}}B \right)-{{\tan }^{2}}B\left( 1+{{\tan }^{2}}A \right)$
Using the distributive property of multiplication over addition, i.e. a(b+c) = ab+ac, we get
LHS $={{\tan }^{2}}A+{{\tan }^{2}}A{{\tan }^{2}}B-{{\tan }^{2}}B-{{\tan }^{2}}A{{\tan }^{2}}B$
Hence, we have
LHS $={{\tan }^{2}}A-{{\tan }^{2}}B$
Hence LHS = RHS.
Hence proved.

Note: Alternatively, we have $\tan x=\dfrac{\sin x}{\cos x}$ and $\sec x=\dfrac{1}{\cos x}$.
Using the above-mentioned formulae, we get
LHS $=\dfrac{{{\sin }^{2}}A}{{{\cos }^{2}}A{{\cos }^{2}}B}-\dfrac{{{\sin }^{2}}B}{{{\cos }^{2}}A{{\cos }^{2}}B}$
Taking ${{\cos }^{2}}A{{\cos }^{2}}B$ as LCM, we get
LHS $=\dfrac{{{\sin }^{2}}A-{{\sin }^{2}}B}{{{\cos }^{2}}A{{\cos }^{2}}B}$
Now, we know that ${{\sin }^{2}}A+{{\cos }^{2}}A=1$
Using, we get
LHS $=\dfrac{{{\sin }^{2}}A\left( {{\sin }^{2}}B+{{\cos }^{2}}B \right)-{{\sin }^{2}}B\left( {{\sin }^{2}}A+{{\cos }^{2}}A \right)}{{{\cos }^{2}}A{{\cos }^{2}}B}$
Using the distributive property of multiplication over addition, we get
LHS $=\dfrac{{{\sin }^{2}}A{{\cos }^{2}}B+{{\sin }^{2}}A{{\sin }^{2}}B-{{\sin }^{2}}A{{\sin }^{2}}B-{{\sin }^{2}}B{{\cos }^{2}}A}{{{\cos }^{2}}A{{\cos }^{2}}B}$
We know that $\dfrac{a+b}{c}=\dfrac{a}{c}+\dfrac{b}{c}$
Using the above identity, we get
LHS $=\dfrac{{{\sin }^{2}}A{{\cos }^{2}}B}{{{\cos }^{2}}A{{\cos }^{2}}B}-\dfrac{{{\sin }^{2}}B{{\cos }^{2}}A}{{{\cos }^{2}}A{{\cos }^{2}}B}$
Hence, we have
LHS $=\dfrac{{{\sin }^{2}}A}{{{\cos }^{2}}A}-\dfrac{{{\sin }^{2}}B}{{{\cos }^{2}}B}$
Using $\tan x=\dfrac{\sin x}{\cos x}$, we get
LHS $={{\tan }^{2}}A-{{\tan }^{2}}B$
Hence LHS = RHS and the given identity is proved.