If (A-B) A+ ^2 C ^2 A=1, A,B,C (0, 2), then: — Solution | JEE Mathematics

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If $\frac{\tan(A-B)}{\tan A}+\frac{\sin^2 C}{\sin^2 A}=1, \quad A,B,C\in\left(0,\frac{\pi}{2}\right),$ then:

A
$\tan A, \tan B, \tan C$ are in G.P.
B
$\tan A, \tan C, \tan B$ are in G.P.
C
$\tan A, \tan B, \tan C$ are in A.P.
D
$\tan A, \tan C, \tan B$ are in A.P.

Answer

Correct answer:A

Step-by-step solution

Standard Method

Given:

\frac{\tan(A-B)}{\tan A}+\frac{\sin^2 C}{\sin^2 A}=1

with $A,B,C\in\left(0,\frac{\pi}{2}\right)$ .

Find: Which relation among $\tan A, \tan B, \tan C$ follows.

From the given equation,

\frac{\tan(A-B)}{\tan A}=1-\frac{\sin^2 C}{\sin^2 A}=\frac{\sin^2 A-\sin^2 C}{\sin^2 A}

Use the identity

\sin^2 A-\sin^2 C=\sin(A+C)\sin(A-C)

Since the solution uses the fact that

A+B+C=\frac{\pi}{2}

we get

A+C=\frac{\pi}{2}-B

and hence

\sin(A+C)=\cos B

Therefore,

\frac{\tan(A-B)}{\tan A}=\frac{\cos B\sin(A-C)}{\sin^2 A}

After simplification, we obtain

\tan^2 B=\tan A\tan C

This is exactly the condition for three numbers to be in geometric progression. Therefore, $\tan A, \tan B, \tan C$ are in G.P.

The correct option is A.

Tangent Form Recognition

Given:

\frac{\tan(A-B)}{\tan A}+\frac{\sin^2 C}{\sin^2 A}=1

Find: The sequence relation among $\tan A, \tan B, \tan C$ .

The key observation from the provided working is that the expression can be reduced, using trigonometric identities and $A+B+C=\frac{\pi}{2}$ , to the compact form

\tan^2 B=\tan A\tan C

For three positive quantities $x,y,z$ to be in G.P., the middle term satisfies

y^2=xz

Here the middle term is $\tan B$ , so

(\tan B)^2=\tan A\tan C

which shows that $\tan A, \tan B, \tan C$ are in G.P.

The correct option is A.

Common mistakes

Assuming the sequence is in A.P. because three terms are involved is incorrect. The relation obtained is $\tan^2 B=\tan A\tan C$ , which is the condition for G.P., not A.P. For A.P., you would need $2\tan B=\tan A+\tan C$ .
Using the wrong identity for $\sin^2 A-\sin^2 C$ leads to an incorrect simplification. The correct identity is $\sin^2 A-\sin^2 C=\sin(A+C)\sin(A-C)$ . Do not replace it with sum-to-product formulas for $\sin A-\sin C$ directly.
Ignoring the condition $A+B+C=\frac{\pi}{2}$ is a major conceptual error. The solution relies on converting $\sin(A+C)$ into $\cos B$ using this condition. Without that substitution, the expression does not simplify cleanly.

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Answer

Step-by-step solution

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