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Question 1: If $\sin \theta + \sin^2 \theta = 1$, prove that $\cos^2 \theta + \cos^4 \theta = 1$.
Solution:
Step 1: Given $\sin \theta = 1 - \sin^2 \theta = \cos^2 \theta$.
Step 2: Substitute $\sin \theta = \cos^2 \theta$ in the expression to be proved.
Step 3: LHS $= \cos^2 \theta + (\cos^2 \theta)^2 = \sin \theta + \sin^2 \theta$.
Step 4: Since $\sin \theta + \sin^2 \theta = 1$, LHS $= 1$. Hence proved. -
Question 2: If $\sec \theta + \tan \theta = p$, show that $\frac{p^2 - 1}{p^2 + 1} = \sin \theta$.
Solution:
Step 1: $p^2 = (\sec \theta + \tan \theta)^2 = \sec^2 \theta + \tan^2 \theta + 2\sec \theta \tan \theta$.
Step 2: $p^2 - 1 = \sec^2 \theta - 1 + \tan^2 \theta + 2\sec \theta \tan \theta = 2\tan^2 \theta + 2\sec \theta \tan \theta = 2\tan \theta(\tan \theta + \sec \theta)$.
Step 3: $p^2 + 1 = \sec^2 \theta + 1 + \tan^2 \theta + 2\sec \theta \tan \theta = 2\sec^2 \theta + 2\sec \theta \tan \theta = 2\sec \theta(\sec \theta + \tan \theta)$.
Step 4: Ratio $= \frac{2\tan \theta(\sec \theta + \tan \theta)}{2\sec \theta(\sec \theta + \tan \theta)} = \frac{\tan \theta}{\sec \theta} = \sin \theta$. -
Question 3: Prove that $\frac{\tan \theta + \sec \theta - 1}{\tan \theta - \sec \theta + 1} = \frac{1 + \sin \theta}{\cos \theta}$.
Solution:
Step 1: Replace $1$ in numerator with $\sec^2 \theta - \tan^2 \theta$.
Step 2: Num $= (\sec \theta + \tan \theta) - (\sec \theta - \tan \theta)(\sec \theta + \tan \theta)$.
Step 3: Factor out $(\sec \theta + \tan \theta)$: Num $= (\sec \theta + \tan \theta)(1 - \sec \theta + \tan \theta)$.
Step 4: Denominator is $\tan \theta - \sec \theta + 1$, which cancels out.
Step 5: Result $= \sec \theta + \tan \theta = \frac{1}{\cos \theta} + \frac{\sin \theta}{\cos \theta} = \frac{1+\sin \theta}{\cos \theta}$. -
Question 4: If $x = a \cos \theta - b \sin \theta$ and $y = a \sin \theta + b \cos \theta$, prove that $x^2 + y^2 = a^2 + b^2$.
Solution:
Step 1: $x^2 = a^2 \cos^2 \theta + b^2 \sin^2 \theta - 2ab \sin \theta \cos \theta$.
Step 2: $y^2 = a^2 \sin^2 \theta + b^2 \cos^2 \theta + 2ab \sin \theta \cos \theta$.
Step 3: Adding both: $x^2 + y^2 = a^2(\cos^2 \theta + \sin^2 \theta) + b^2(\sin^2 \theta + \cos^2 \theta)$.
Step 4: $x^2 + y^2 = a^2(1) + b^2(1) = a^2 + b^2$. -
Question 5: If $\tan A = n \tan B$ and $\sin A = m \sin B$, prove that $\cos^2 A = \frac{m^2 - 1}{n^2 - 1}$.
Solution:
Step 1: From given, $\cot B = \frac{n}{\tan A}$ and $\csc B = \frac{m}{\sin A}$.
Step 2: Use identity $\csc^2 B - \cot^2 B = 1$.
Step 3: $\frac{m^2}{\sin^2 A} - \frac{n^2}{\tan^2 A} = 1 \Rightarrow \frac{m^2}{\sin^2 A} - \frac{n^2 \cos^2 A}{\sin^2 A} = 1$.
Step 4: $m^2 - n^2 \cos^2 A = \sin^2 A = 1 - \cos^2 A$.
Step 5: $m^2 - 1 = \cos^2 A (n^2 - 1) \Rightarrow \cos^2 A = \frac{m^2 - 1}{n^2 - 1}$. -
Question 6: Prove that $(1 + \cot A - \csc A)(1 + \tan A + \sec A) = 2$.
Solution:
Step 1: Convert to sine and cosine: $(\frac{\sin A + \cos A - 1}{\sin A}) (\frac{\cos A + \sin A + 1}{\cos A})$.
Step 2: Numerator is of form $(x-1)(x+1) = x^2 - 1$ where $x = \sin A + \cos A$.
Step 3: $(\sin A + \cos A)^2 - 1 = \sin^2 A + \cos^2 A + 2\sin A \cos A - 1 = 1 + 2\sin A \cos A - 1 = 2\sin A \cos A$.
Step 4: Divide by denominator $\sin A \cos A$: $\frac{2\sin A \cos A}{\sin A \cos A} = 2$. -
Question 7: If $\cos \theta + \sin \theta = \sqrt{2} \cos \theta$, show that $\cos \theta - \sin \theta = \sqrt{2} \sin \theta$.
Solution:
Step 1: Square the given equation: $\cos^2 \theta + \sin^2 \theta + 2\sin \theta \cos \theta = 2\cos^2 \theta$.
Step 2: $1 + 2\sin \theta \cos \theta = 2\cos^2 \theta \Rightarrow 2\sin \theta \cos \theta = 2\cos^2 \theta - 1$.
Step 3: Consider $(\cos \theta - \sin \theta)^2 = \cos^2 \theta + \sin^2 \theta - 2\sin \theta \cos \theta = 1 - (2\cos^2 \theta - 1)$.
Step 4: $= 2 - 2\cos^2 \theta = 2(1 - \cos^2 \theta) = 2\sin^2 \theta$.
Step 5: Taking square root: $\cos \theta - \sin \theta = \sqrt{2} \sin \theta$. -
Question 8: Prove that $\frac{\cos A}{1 - \tan A} + \frac{\sin A}{1 - \cot A} = \sin A + \cos A$.
Solution:
Step 1: Convert tan/cot to sin/cos: $\frac{\cos^2 A}{\cos A - \sin A} + \frac{\sin^2 A}{\sin A - \cos A}$.
Step 2: Change sign of second term: $\frac{\cos^2 A}{\cos A - \sin A} - \frac{\sin^2 A}{\cos A - \sin A}$.
Step 3: Combine: $\frac{\cos^2 A - \sin^2 A}{\cos A - \sin A} = \frac{(\cos A - \sin A)(\cos A + \sin A)}{\cos A - \sin A}$.
Step 4: Result is $\cos A + \sin A$. -
Question 9: If $a \cos \theta + b \sin \theta = m$ and $a \sin \theta - b \cos \theta = n$, prove that $a^2 + b^2 = m^2 + n^2$.
Solution:
Step 1: Calculate $m^2 + n^2$.
Step 2: $(a \cos \theta + b \sin \theta)^2 + (a \sin \theta - b \cos \theta)^2$.
Step 3: Expand: $(a^2 \cos^2 \theta + b^2 \sin^2 \theta + 2ab \dots) + (a^2 \sin^2 \theta + b^2 \cos^2 \theta - 2ab \dots)$.
Step 4: Group terms: $a^2(\cos^2 \theta + \sin^2 \theta) + b^2(\sin^2 \theta + \cos^2 \theta) = a^2 + b^2$. -
Question 10: Prove that $\sqrt{\frac{\sec \theta - 1}{\sec \theta + 1}} + \sqrt{\frac{\sec \theta + 1}{\sec \theta - 1}} = 2 \csc \theta$.
Solution:
Step 1: Cross multiply in numerator: $\frac{(\sec \theta - 1) + (\sec \theta + 1)}{\sqrt{\sec^2 \theta - 1}}$.
Step 2: Numerator $= 2\sec \theta$. Denominator $= \sqrt{\tan^2 \theta} = \tan \theta$.
Step 3: $\frac{2\sec \theta}{\tan \theta} = \frac{2/\cos \theta}{\sin \theta/\cos \theta} = \frac{2}{\sin \theta} = 2\csc \theta$.