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Q3(iii):

Choose the correct option. Justify your choice. (iii) $(\sec A + \tan A) (1 – \sin A) =$

Solution :

Given: An expression $(\sec A + \tan A)(1 - \sin A)$.

To find: The simplified value of the given expression by choosing the correct option among the standard trigonometric identities.

Step 1: Expressing trigonometric ratios in terms of sine and cosine.

We know the fundamental definitions of trigonometric ratios:

$\sec A = \frac{1}{\cos A}$ [Since secant is the reciprocal of cosine]

$\tan A = \frac{\sin A}{\cos A}$ [Since tangent is the ratio of sine to cosine]

Step 2: Substituting these values into the given expression.

Let the expression be $E$.

$E = \left( \frac{1}{\cos A} + \frac{\sin A}{\cos A} \right) (1 - \sin A)$

Step 3: Simplifying the term inside the parentheses.

Since the denominators are the same, we can combine the fractions:

$E = \left( \frac{1 + \sin A}{\cos A} \right) (1 - \sin A)$

Step 4: Performing the multiplication.

Multiply the numerators together:

$E = \frac{(1 + \sin A)(1 - \sin A)}{\cos A}$

Step 5: Applying the algebraic identity $(a + b)(a - b) = a^2 - b^2$.

Here, $a = 1$ and $b = \sin A$.

$E = \frac{1^2 - \sin^2 A}{\cos A}$

$E = \frac{1 - \sin^2 A}{\cos A}$

Step 6: Applying the Pythagorean identity.

We know the identity: $\sin^2 A + \cos^2 A = 1$.

Rearranging this gives: $1 - \sin^2 A = \cos^2 A$.

Substituting this into our expression:

$E = \frac{\cos^2 A}{\cos A}$

Step 7: Final simplification.

$E = \frac{\cos A \cdot \cos A}{\cos A}$

$E = \cos A$ [By canceling the common factor $\cos A$ in the numerator and denominator]

Final Answer: The simplified value of the expression is $\cos A$.


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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