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Math Notes For Class 11 Chapter 7 Ex 7.1 Solved & Quiz
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Math Notes For Class 11 Chapter 7 Ex 7.1 Solved & Quiz
Trigonometry forms the structural bedrock of modern calculus, physics, and engineering analysis. An identity is an equation that holds true for all admissible values of the domain variable. Exercise 7.1 focuses on the Fundamental Pythagorean Identities and the Fundamental Distance/Sum & Difference Formulas.
| Identity Type | Mathematical Formula | Primary Application Domain |
|---|---|---|
| Pythagorean I | $$\sin^2\theta + \cos^2\theta = 1$$ | Converting between sine and cosine terms |
| Pythagorean II | $$1 + \tan^2\theta = \sec^2\theta$$ | Simplifying expressions involving secant and tangent |
| Pythagorean III | $$1 + \cot^2\theta = \csc^2\theta$$ | Simplifying cosecant and cotangent forms |
| Cosine of Difference | $$\cos(\alpha - \beta) = \cos\alpha\cos\beta + \sin\alpha\sin\beta$$ | Distance formula proof on unit circle |
| Sine of Sum | $$\sin(\alpha + \beta) = \sin\alpha\cos\beta + \cos\alpha\sin\beta$$ | Angle addition in non-standard evaluations |
| Tangent of Difference | $$\tan(\alpha - \beta) = \frac{\tan\alpha - \tan\beta}{1 + \tan\alpha\tan\beta}$$ | Evaluating slopes and angle transformations |
Question 1 (Part i): Simplify the algebraic trigonometric expression: $\frac{\sin\theta + \tan\theta}{1 + \cos\theta}$.
Solution:
Step 1: Express $\tan\theta$ in terms of $\sin\theta$ and $\cos\theta$:
$$\text{Numerator} = \sin\theta + \frac{\sin\theta}{\cos\theta}$$
Step 2: Factor out $\sin\theta$ from the numerator:
$$\text{Numerator} = \sin\theta \left(1 + \frac{1}{\cos\theta}\right) = \sin\theta \left(\frac{\cos\theta + 1}{\cos\theta}\right)$$
Step 3: Substitute the rewritten numerator into the original rational expression:
$$\frac{\sin\theta \left(\frac{1 + \cos\theta}{\cos\theta}\right)}{1 + \cos\theta}$$
Step 4: Cancel the common factor $(1 + \cos\theta)$ present in both numerator and denominator (assuming $\cos\theta \neq -1$):
$$= \frac{\sin\theta}{\cos\theta} = \tan\theta$$
Answer: $$\tan\theta$$
Question 1 (Part ii): Simplify the expression: $\frac{\cos\theta}{1 - \sin\theta} - \frac{\cos\theta}{1 + \sin\theta}$.
Solution:
Step 1: Take the Least Common Multiple (LCM) of the denominators, which is $(1 - \sin\theta)(1 + \sin\theta)$:
$$\text{LCM} = 1 - \sin^2\theta = \cos^2\theta$$
Step 2: Combine the fractions over the common denominator:
$$E = \frac{\cos\theta(1 + \sin\theta) - \cos\theta(1 - \sin\theta)}{(1 - \sin\theta)(1 + \sin\theta)}$$
Step 3: Expand the numerator:
$$\text{Numerator} = \cos\theta + \cos\theta\sin\theta - \cos\theta + \cos\theta\sin\theta = 2\cos\theta\sin\theta$$
Step 4: Divide by the simplified denominator $\cos^2\theta$:
$$E = \frac{2\cos\theta\sin\theta}{\cos^2\theta} = 2 \cdot \frac{\sin\theta}{\cos\theta} = 2\tan\theta$$
Answer: $$2\tan\theta$$
Question 2: Prove the trigonometric identity: $\frac{1 - \sin\theta}{\cos\theta} = \frac{\cos\theta}{1 + \sin\theta}$.
Solution:
Step 1: Start with the Left-Hand Side (LHS):
$$\text{LHS} = \frac{1 - \sin\theta}{\cos\theta}$$
Step 2: Multiply the numerator and denominator by the conjugate of the numerator, $(1 + \sin\theta)$:
$$\text{LHS} = \frac{(1 - \sin\theta)(1 + \sin\theta)}{\cos\theta(1 + \sin\theta)}$$
Step 3: Apply the difference of squares identity $(a-b)(a+b) = a^2 - b^2$ in the numerator:
$$\text{Numerator} = 1 - \sin^2\theta = \cos^2\theta$$
Step 4: Substitute back into the expression and simplify:
$$\text{LHS} = \frac{\cos^2\theta}{\cos\theta(1 + \sin\theta)} = \frac{\cos\theta}{1 + \sin\theta} = \text{RHS}$$
Since LHS = RHS, the identity is verified.
Answer: $$\frac{\cos\theta}{1 + \sin\theta}$$
Question 3: Prove that $(\tan\theta + \cot\theta)^2 = \sec^2\theta + \csc^2\theta$.
Solution:
Step 1: Consider the Left-Hand Side (LHS) and convert to sine and cosine:
$$\text{LHS} = \left(\frac{\sin\theta}{\cos\theta} + \frac{\cos\theta}{\sin\theta}\right)^2$$
Step 2: Combine the terms inside the parentheses using LCM:
$$\frac{\sin\theta}{\cos\theta} + \frac{\cos\theta}{\sin\theta} = \frac{\sin^2\theta + \cos^2\theta}{\sin\theta\cos\theta} = \frac{1}{\sin\theta\cos\theta}$$
Step 3: Square the result:
$$\text{LHS} = \left(\frac{1}{\sin\theta\cos\theta}\right)^2 = \frac{1}{\sin^2\theta\cos^2\theta}$$
Step 4: Consider the Right-Hand Side (RHS) and express in sine and cosine:
$$\text{RHS} = \sec^2\theta + \csc^2\theta = \frac{1}{\cos^2\theta} + \frac{1}{\sin^2\theta}$$
Step 5: Take the LCM of the RHS:
$$\text{RHS} = \frac{\sin^2\theta + \cos^2\theta}{\cos^2\theta\sin^2\theta} = \frac{1}{\sin^2\theta\cos^2\theta}$$
Since $\text{LHS} = \text{RHS} = \frac{1}{\sin^2\theta\cos^2\theta}$, the identity is proven.
Answer: $$\sec^2\theta + \csc^2\theta$$
Question 4 (Part i): Find the exact value of $\sin(105^\circ)$ without using a calculator.
Solution:
Step 1: Decompose $105^\circ$ into standard known angles: $105^\circ = 60^\circ + 45^\circ$.
Step 2: Apply the sine sum formula $\sin(\alpha + \beta) = \sin\alpha\cos\beta + \cos\alpha\sin\beta$:
$$\sin(105^\circ) = \sin(60^\circ + 45^\circ) = \sin(60^\circ)\cos(45^\circ) + \cos(60^\circ)\sin(45^\circ)$$
Step 3: Substitute exact values ($\sin 60^\circ = \frac{\sqrt{3}}{2}$, $\cos 45^\circ = \frac{\sqrt{2}}{2}$, $\cos 60^\circ = \frac{1}{2}$, $\sin 45^\circ = \frac{\sqrt{2}}{2}$):
$$\sin(105^\circ) = \left(\frac{\sqrt{3}}{2}\right)\left(\frac{\sqrt{2}}{2}\right) + \left(\frac{1}{2}\right)\left(\frac{\sqrt{2}}{2}\right)$$
Step 4: Multiply and combine fractions:
$$\sin(105^\circ) = \frac{\sqrt{6}}{4} + \frac{\sqrt{2}}{4} = \frac{\sqrt{6} + \sqrt{2}}{4}$$
Answer: $$\frac{\sqrt{6} + \sqrt{2}}{4}$$
Question 4 (Part ii): Find the exact value of $\cos(75^\circ)$ without using a calculator.
Solution:
Step 1: Express $75^\circ$ as $45^\circ + 30^\circ$.
Step 2: Apply the cosine sum formula $\cos(\alpha + \beta) = \cos\alpha\cos\beta - \sin\alpha\sin\beta$:
$$\cos(75^\circ) = \cos(45^\circ + 30^\circ) = \cos(45^\circ)\cos(30^\circ) - \sin(45^\circ)\sin(30^\circ)$$
Step 3: Substitute exact values ($\cos 45^\circ = \frac{\sqrt{2}}{2}$, $\cos 30^\circ = \frac{\sqrt{3}}{2}$, $\sin 45^\circ = \frac{\sqrt{2}}{2}$, $\sin 30^\circ = \frac{1}{2}$):
$$\cos(75^\circ) = \left(\frac{\sqrt{2}}{2}\right)\left(\frac{\sqrt{3}}{2}\right) - \left(\frac{\sqrt{2}}{2}\right)\left(\frac{1}{2}\right)$$
Step 4: Combine terms:
$$\cos(75^\circ) = \frac{\sqrt{6}}{4} - \frac{\sqrt{2}}{4} = \frac{\sqrt{6} - \sqrt{2}}{4}$$
Answer: $$\frac{\sqrt{6} - \sqrt{2}}{4}$$
Question 5: Prove that $\frac{\cos 11^\circ + \sin 11^\circ}{\cos 11^\circ - \sin 11^\circ} = \tan 56^\circ$.
Solution:
Step 1: Consider the Left-Hand Side (LHS) and divide both numerator and denominator by $\cos 11^\circ$:
$$\text{LHS} = \frac{\frac{\cos 11^\circ}{\cos 11^\circ} + \frac{\sin 11^\circ}{\cos 11^\circ}}{\frac{\cos 11^\circ}{\cos 11^\circ} - \frac{\sin 11^\circ}{\cos 11^\circ}} = \frac{1 + \tan 11^\circ}{1 - \tan 11^\circ}$$
Step 2: Recall that $\tan 45^\circ = 1$. Rewrite the constant $1$ as $\tan 45^\circ$:
$$\text{LHS} = \frac{\tan 45^\circ + \tan 11^\circ}{1 - (\tan 45^\circ)(\tan 11^\circ)}$$
Step 3: Recognize this as the expansion of $\tan(\alpha + \beta) = \frac{\tan\alpha + \tan\beta}{1 - \tan\alpha\tan\beta}$ with $\alpha = 45^\circ$ and $\beta = 11^\circ$:
$$\text{LHS} = \tan(45^\circ + 11^\circ) = \tan 56^\circ = \text{RHS}$$
Answer: $$\tan 56^\circ$$
Question 6: Prove the formula identity $\frac{\sin(\alpha+\beta)}{\cos\alpha \cos\beta} = \tan\alpha + \tan\beta$.
Solution:
Step 1: Start with the Left-Hand Side (LHS) and expand $\sin(\alpha+\beta)$:
$$\text{LHS} = \frac{\sin\alpha\cos\beta + \cos\alpha\sin\beta}{\cos\alpha\cos\beta}$$
Step 2: Split the single fraction into two separate fractions over the common denominator:
$$\text{LHS} = \frac{\sin\alpha\cos\beta}{\cos\alpha\cos\beta} + \frac{\cos\alpha\sin\beta}{\cos\alpha\cos\beta}$$
Step 3: Cancel $\cos\beta$ in the first term and $\cos\alpha$ in the second term:
$$\text{LHS} = \frac{\sin\alpha}{\cos\alpha} + \frac{\sin\beta}{\cos\beta}$$
Step 4: Use the quotient identity $\tan\theta = \frac{\sin\theta}{\cos\theta}$:
$$\text{LHS} = \tan\alpha + \tan\beta = \text{RHS}$$
Answer: $$\tan\alpha + \tan\beta$$
Q1: What is the simplified value of $\sin^2(30^\circ) + \cos^2(30^\circ)$?
Q2: Express $\cos(\alpha - \beta)$ in fundamental expanded form:
Q3: The expression $\sec^2\theta - \tan^2\theta$ is identical to: