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Q3(i):

Choose the correct option. Justify your choice. (i) $9 \sec^2 A – 9 \tan^2 A =$

Solution :

Given: An algebraic expression involving trigonometric functions: $9 \sec^2 A - 9 \tan^2 A$.

To Find: The numerical value of the expression by simplifying it using trigonometric identities.

Step 1: Factoring the expression
The given expression is $9 \sec^2 A - 9 \tan^2 A$. We observe that the constant $9$ is common to both terms. We can factor it out using the distributive property of multiplication over subtraction:
$9 \sec^2 A - 9 \tan^2 A = 9(\sec^2 A - \tan^2 A)$

Step 2: Identifying the relevant trigonometric identity
We recall the fundamental trigonometric identity relating the secant and tangent functions:
$1 + \tan^2 A = \sec^2 A$
[Since the identity holds for all values of $A$ where the functions are defined]

Step 3: Rearranging the identity
To match the expression inside the parentheses, we rearrange the identity $1 + \tan^2 A = \sec^2 A$ by subtracting $\tan^2 A$ from both sides:
$1 = \sec^2 A - \tan^2 A$
[By the subtraction property of equality]

Step 4: Substitution and Final Calculation
Now, substitute the value of $(\sec^2 A - \tan^2 A)$ into the factored expression from Step 1:
$9(\sec^2 A - \tan^2 A) = 9(1)$
[Substituting $1$ for the expression $\sec^2 A - \tan^2 A$]
$9(1) = 9$

Justification: The expression simplifies to $9$ because the difference between the square of the secant and the square of the tangent of an angle is always equal to $1$, as derived from the Pythagorean identity $1 + \tan^2 A = \sec^2 A$.

Final Answer: 9


More Questions from Class 10 Mathematics Introduction to Trigonometry EXERCISE 8.3


CBSE Solutions for Class 10 Mathematics Introduction to Trigonometry


Chapters in CBSE - Class 10 Mathematics


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