# Introduction to Trigonometry - Class 10 - Mathematics

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## Extra Questions - Introduction to Trigonometry | NCERT | Mathematics | Class 10

If $a \sin \theta + b \cos \theta = c$, then prove that $a \cos \theta - b \sin \theta = \sqrt{a^{2} + b^{2} - c^{2}}$.

Given the equation: $$ a \sin \theta + b \cos \theta = c $$

We need to prove that: $$ a \cos \theta - b \sin \theta = \sqrt{a^2 + b^2 - c^2} $$

**Step 1:** Square the given equation on both sides:
$$
(a \sin \theta + b \cos \theta)^2 = c^2
$$

Expanding using the identity $(x + y)^2 = x^2 + 2xy + y^2$: $$ a^2 \sin^2 \theta + 2ab \sin \theta \cos \theta + b^2 \cos^2 \theta = c^2 $$

**Step 2:** Use the Pythagorean identity $\sin^2 \theta + \cos^2 \theta = 1$:
Replace $\sin^2 \theta$ by $1 - \cos^2 \theta$ and $\cos^2 \theta$ by $1 - \sin^2 \theta$:
$$
a^2 (1 - \cos^2 \theta) + 2ab \sin \theta \cos \theta + b^2 (1 - \sin^2 \theta) = c^2
$$
This simplifies to:
$$
a^2 + b^2 - a^2 \cos^2 \theta - b^2 \sin^2 \theta + 2ab \sin \theta \cos \theta = c^2
$$

Rearranging and using $a^2 + b^2 - c^2$ on the left-hand side: $$ a^2 + b^2 - c^2 = a^2 \cos^2 \theta + b^2 \sin^2 \theta - 2ab \sin \theta \cos \theta $$

Substituting back in terms of sine and cosine: $$ a^2 + b^2 - c^2 = (a \cos \theta - b \sin \theta)^2 $$

**Step 3:** Taking the square root on both sides:
$$
\sqrt{a^2 + b^2 - c^2} = |a \cos \theta - b \sin \theta|
$$

Since trigonometric functions could have either sign based on the quadrant and the equality must hold for all $\theta$, we can drop the absolute value under the presumption of the correct sign handling or context: $$ a \cos \theta - b \sin \theta = \sqrt{a^2 + b^2 - c^2} $$

This final equation matches what we aimed to prove, thus validating our formulation and manipulation of trigonometric identities and algebraic expressions.

If $\sin \theta + \cos \theta = 1.2$, then $\sin \theta \cdot \cos \theta =$

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