CBSE Class 10 Mathematics Trigonometry Assignment Set 11

Question. Express the other t-ratios of \(\angle A\) in terms of
(a) \(\sin A\)
(b) \(\cos A\)
(c) \(\tan A\)
(d) \(\cot A\)
(e) \(\sec A\)
(f) \(\csc A\)
Answer:
(a) In terms of \(\sin A\):
\(\cos A = \sqrt{1 - \sin^2 A}\)
\(\tan A = \frac{\sin A}{\sqrt{1 - \sin^2 A}}\)
\(\cot A = \frac{\sqrt{1 - \sin^2 A}}{\sin A}\)
\(\sec A = \frac{1}{\sqrt{1 - \sin^2 A}}\)
\(\csc A = \frac{1}{\sin A}\)

(b) In terms of \(\cos A\):
\(\sin A = \sqrt{1 - \cos^2 A}\)
\(\tan A = \frac{\sqrt{1 - \cos^2 A}}{\cos A}\)
\(\cot A = \frac{\cos A}{\sqrt{1 - \cos^2 A}}\)
\(\sec A = \frac{1}{\cos A}\)
\(\csc A = \frac{1}{\sqrt{1 - \cos^2 A}}\)

(c) In terms of \(\tan A\):
\(\sin A = \frac{\tan A}{\sqrt{1 + \tan^2 A}}\)
\(\cos A = \frac{1}{\sqrt{1 + \tan^2 A}}\)
\(\cot A = \frac{1}{\tan A}\)
\(\sec A = \sqrt{1 + \tan^2 A}\)
\(\csc A = \frac{\sqrt{1 + \tan^2 A}}{\tan A}\)

(d) In terms of \(\cot A\):
\(\sin A = \frac{1}{\sqrt{1 + \cot^2 A}}\)
\(\cos A = \frac{\cot A}{\sqrt{1 + \cot^2 A}}\)
\(\tan A = \frac{1}{\cot A}\)
\(\sec A = \frac{\sqrt{1 + \cot^2 A}}{\cot A}\)
\(\csc A = \sqrt{1 + \cot^2 A}\)

(e) In terms of \(\sec A\):
\(\sin A = \frac{\sqrt{\sec^2 A - 1}}{\sec A}\)
\(\cos A = \frac{1}{\sec A}\)
\(\tan A = \sqrt{\sec^2 A - 1}\)
\(\cot A = \frac{1}{\sqrt{\sec^2 A - 1}}\)
\(\csc A = \frac{\sec A}{\sqrt{\sec^2 A - 1}}\)

(f) In terms of \(\csc A\):
\(\sin A = \frac{1}{\csc A}\)
\(\cos A = \frac{\sqrt{\csc^2 A - 1}}{\csc A}\)
\(\tan A = \frac{1}{\sqrt{\csc^2 A - 1}}\)
\(\cot A = \sqrt{\csc^2 A - 1}\)
\(\sec A = \frac{\csc A}{\sqrt{\csc^2 A - 1}}\)
In simple words: We can write any trigonometric ratio using any other ratio by applying basic algebraic identities like \(\sin^2 A + \cos^2 A = 1\) and reciprocal relations.
Exam Tip: Memorize the three fundamental identities: \(\sin^2 A + \cos^2 A = 1\), \(1 + \tan^2 A = \sec^2 A\), and \(1 + \cot^2 A = \csc^2 A\) to make these transformations easier.

Question. Prove \(\frac{\sin^2 44^{\circ} + \sin^2 46^{\circ}}{\cos^2 47^{\circ} + \cos^2 43^{\circ}} = 1\)
Answer:
Let us simplify the left hand side (LHS) of the given expression:
\(\text{LHS} = \frac{\sin^2 44^{\circ} + \sin^2 46^{\circ}}{\cos^2 47^{\circ} + \cos^2 43^{\circ}}\)
Using complementary angle relations, we know that:
\(\sin 46^{\circ} = \sin(90^{\circ} - 44^{\circ}) = \cos 44^{\circ}\)
And similarly for the denominator:
\(\cos 43^{\circ} = \cos(90^{\circ} - 47^{\circ}) = \sin 47^{\circ}\)
By substituting these values back into the expression, we get:
\(\text{LHS} = \frac{\sin^2 44^{\circ} + \cos^2 44^{\circ}}{\cos^2 47^{\circ} + \sin^2 47^{\circ}}\)
We apply the standard trigonometric identity \(\sin^2 \theta + \cos^2 \theta = 1\):
\(\text{LHS} = \frac{1}{1} = 1 = \text{RHS}\)
Hence, the identity is proved.
In simple words: Change the angles using complementary formulas so that both the top and bottom expressions simplify to 1 using \(\sin^2 \theta + \cos^2 \theta = 1\).

Exam Tip: Identify complementary pairs (like 44 and 46, or 47 and 43) that add up to 90 degrees, and convert one of them to simplify the expression.

Question. Prove \(\frac{\sin(90^{\circ}-A)\sin A}{\tan A} = 1 - \sin^2 A\)
Answer:
Starting with the left hand side (LHS):
\(\text{LHS} = \frac{\sin(90^{\circ}-A)\sin A}{\tan A}\)
We can substitute \(\sin(90^{\circ}-A) = \cos A\) and \(\tan A = \frac{\sin A}{\cos A}\):
\(\text{LHS} = \frac{\cos A \sin A}{\frac{\sin A}{\cos A}}\)

\(\implies \text{LHS} = \cos A \sin A \times \frac{\cos A}{\sin A}\)

\(\implies \text{LHS} = \cos^2 A\)
Using the identity \(\cos^2 A = 1 - \sin^2 A\):
\(\text{LHS} = 1 - \sin^2 A = \text{RHS}\)
Hence, proved.
In simple words: Turn \(\sin(90^{\circ}-A)\) into \(\cos A\) and \(\tan A\) into \(\sin A / \cos A\). After canceling out \(\sin A\), you are left with \(\cos^2 A\), which is equal to \(1 - \sin^2 A\).

Exam Tip: Expressing all terms in basic forms like sine and cosine is a reliable way to solve trigonometric proofs.

Question. Prove \(\sin(90^{\circ}-\theta)\cos(90^{\circ}-\theta) = \frac{\tan\theta}{1+\tan^2\theta}\)
Answer:
Let us simplify the left hand side (LHS) first:
\(\text{LHS} = \sin(90^{\circ}-\theta)\cos(90^{\circ}-\theta)\)
We apply the complementary angle relations \(\sin(90^{\circ}-\theta) = \cos\theta\) and \(\cos(90^{\circ}-\theta) = \sin\theta\):
\(\text{LHS} = \cos\theta \sin\theta\)

Now, let us simplify the right hand side (RHS):
\(\text{RHS} = \frac{\tan\theta}{1+\tan^2\theta}\)
Using the identity \(1+\tan^2\theta = \sec^2\theta\):
\(\text{RHS} = \frac{\tan\theta}{\sec^2\theta}\)
Convert both terms into sine and cosine:
\(\text{RHS} = \frac{\frac{\sin\theta}{\cos\theta}}{\frac{1}{\cos^2\theta}}\)

\(\implies \text{RHS} = \frac{\sin\theta}{\cos\theta} \times \cos^2\theta\)

\(\implies \text{RHS} = \sin\theta \cos\theta\)
Since \(\text{LHS} = \text{RHS}\), the identity is verified.
In simple words: Simplify both sides separately. The left side becomes \(\cos\theta \sin\theta\), and the right side also simplifies to \(\sin\theta \cos\theta\) after converting to sine and cosine.

Exam Tip: When both sides look slightly complex, simplifying LHS and RHS separately to the same expression is a perfectly valid method.

Question. Prove \(\sqrt{\frac{1+\sin A}{1-\sin A}} = \sec A + \tan A\)
Answer:
Starting with the left hand side (LHS):
\(\text{LHS} = \sqrt{\frac{1+\sin A}{1-\sin A}}\)
To eliminate the fraction in the square root, we multiply the numerator and denominator by the conjugate of the denominator, which is \((1+\sin A)\):
\(\text{LHS} = \sqrt{\frac{(1+\sin A)(1+\sin A)}{(1-\sin A)(1+\sin A)}}\)

\(\implies \text{LHS} = \sqrt{\frac{(1+\sin A)^2}{1-\sin^2 A}}\)
Since \(1-\sin^2 A = \cos^2 A\), we can substitute this:
\(\text{LHS} = \sqrt{\frac{(1+\sin A)^2}{\cos^2 A}}\)
Taking the square root gives:
\(\text{LHS} = \frac{1+\sin A}{\cos A}\)
Splitting the fraction into two parts:
\(\text{LHS} = \frac{1}{\cos A} + \frac{\sin A}{\cos A}\)

\(\implies \text{LHS} = \sec A + \tan A = \text{RHS}\)
Hence, proved.
In simple words: Multiply the top and bottom inside the square root by \((1+\sin A)\) to make the bottom \(\cos^2 A\). Once you remove the square root, split the fraction into two simple parts to get the answer.

Exam Tip: Multiplying by the conjugate is a standard technique when you see expressions like \(\sqrt{\frac{1 \pm \sin\theta}{1 \mp \sin\theta}}\) under a radical sign.

Question. Prove \(\sec A + \tan A = \frac{1}{\sec A - \tan A}\)
Answer:
Let us start with the right hand side (RHS):
\(\text{RHS} = \frac{1}{\sec A - \tan A}\)
Multiply the numerator and the denominator by \(\sec A + \tan A\):
\(\text{RHS} = \frac{1 \times (\sec A + \tan A)}{(\sec A - \tan A)(\sec A + \tan A)}\)

\(\implies \text{RHS} = \frac{\sec A + \tan A}{\sec^2 A - \tan^2 A}\)
We know the identity \(\sec^2 A - \tan^2 A = 1\). Substituting this gives:
\(\text{RHS} = \frac{\sec A + \tan A}{1} = \sec A + \tan A = \text{LHS}\)
Hence, the statement is proved.
In simple words: Multiply the top and bottom of the fraction by \(\sec A + \tan A\). The bottom becomes \(\sec^2 A - \tan^2 A\), which is just 1, leaving you with the desired expression.

Exam Tip: Remember that \(\sec^2 \theta - \tan^2 \theta = 1\) means \(\sec\theta - \tan\theta\) and \(\sec\theta + \tan\theta\) are reciprocals of each other.

Question. Prove \(\frac{1+\sin A}{\cos A} = \frac{\cos A}{1-\sin A}\)
Answer:
Starting with the left hand side (LHS):
\(\text{LHS} = \frac{1+\sin A}{\cos A}\)
Multiply both the numerator and denominator by \((1-\sin A)\):
\(\text{LHS} = \frac{(1+\sin A)(1-\sin A)}{\cos A(1-\sin A)}\)

\(\implies \text{LHS} = \frac{1-\sin^2 A}{\cos A(1-\sin A)}\)
Using \(1-\sin^2 A = \cos^2 A\):
\(\text{LHS} = \frac{\cos^2 A}{\cos A(1-\sin A)}\)
Cancel one \(\cos A\) from the numerator and denominator:
\(\text{LHS} = \frac{\cos A}{1-\sin A} = \text{RHS}\)
Hence, proved.
In simple words: Multiply top and bottom by \((1-\sin A)\). The top simplifies to \(\cos^2 A\), allowing you to cancel one \(\cos A\) and get the right-hand side.

Exam Tip: This type of fraction conversion is useful for simplifying larger trigonometric fractions in more complex problems.

Question. Prove \(\csc A - \cot A = \frac{1}{\csc A + \cot A}\)
Answer:
Let us begin with the right hand side (RHS):
\(\text{RHS} = \frac{1}{\csc A + \cot A}\)
Multiply the numerator and denominator by \(\csc A - \cot A\):
\(\text{RHS} = \frac{\csc A - \cot A}{(\csc A + \cot A)(\csc A - \cot A)}\)

\(\implies \text{RHS} = \frac{\csc A - \cot A}{\csc^2 A - \cot^2 A}\)
Using the identity \(\csc^2 A - \cot^2 A = 1\):
\(\text{RHS} = \frac{\csc A - \cot A}{1} = \csc A - \cot A = \text{LHS}\)
Hence, proved.
In simple words: Multiply the top and bottom of the fraction by \(\csc A - \cot A\). The denominator simplifies to 1, leaving you with the left hand side.

Exam Tip: Just like secant and tangent, cosecant and cotangent are related by the identity \(\csc^2 A - \cot^2 A = 1\), meaning they are reciprocal conjugates.

Question. Prove \(\frac{\cos A}{1+\sin A} + \frac{1+\sin A}{\cos A} = \frac{2}{\cos A}\)
Answer:
Let us combine the fractions on the left hand side (LHS) using a common denominator:
\(\text{LHS} = \frac{\cos^2 A + (1+\sin A)^2}{\cos A(1+\sin A)}\)
Expand the term \((1+\sin A)^2\):
\(\text{LHS} = \frac{\cos^2 A + 1 + 2\sin A + \sin^2 A}{\cos A(1+\sin A)}\)
Group \(\sin^2 A + \cos^2 A\) together, which equals 1:
\(\text{LHS} = \frac{(\sin^2 A + \cos^2 A) + 1 + 2\sin A}{\cos A(1+\sin A)}\)

\(\implies \text{LHS} = \frac{1 + 1 + 2\sin A}{\cos A(1+\sin A)}\)

\(\implies \text{LHS} = \frac{2 + 2\sin A}{\cos A(1+\sin A)}\)
Factor out 2 from the numerator:
\(\text{LHS} = \frac{2(1+\sin A)}{\cos A(1+\sin A)}\)
Cancel the common term \((1+\sin A)\):
\(\text{LHS} = \frac{2}{\cos A} = \text{RHS}\)
Hence, the identity is proved.
In simple words: Add the fractions together by finding a common denominator. Simplify the numerator using \(\sin^2 A + \cos^2 A = 1\), factor out 2, and cancel the common bracket.

Exam Tip: Be careful when expanding \((1+\sin A)^2\); make sure not to forget the middle term \(2\sin A\).

Question. Prove \(\sec^4 A - \tan^4 A = \sec^2 A + \tan^2 A\)
Answer:
Let us work with the left hand side (LHS):
\(\text{LHS} = \sec^4 A - \tan^4 A\)
This is a difference of squares, so we can factorize it:
\(\text{LHS} = (\sec^2 A)^2 - (\tan^2 A)^2\)

\(\implies \text{LHS} = (\sec^2 A - \tan^2 A)(\sec^2 A + \tan^2 A)\)
Applying the identity \(\sec^2 A - \tan^2 A = 1\):
\(\text{LHS} = 1 \times (\sec^2 A + \tan^2 A)\)

\(\implies \text{LHS} = \sec^2 A + \tan^2 A = \text{RHS}\)
Hence, proved.
In simple words: Factor the difference of fourth powers into a product of squares, like \((a^2-b^2)(a^2+b^2)\). Since the minus part equals 1, only the plus part remains.

Exam Tip: Algebraic factorization rules like \(a^2 - b^2 = (a-b)(a+b)\) are highly useful when solving high-power trigonometric identities.

Question. If \(\sin\theta + \cos\theta = 1\) Prove \(\sin\theta \cos\theta = 0\)
Answer:
We are given the equation:
\(\sin\theta + \cos\theta = 1\)
Let us square both sides of this equation:
\((\sin\theta + \cos\theta)^2 = 1^2\)
Expanding the left side yields:
\(\sin^2\theta + \cos^2\theta + 2\sin\theta\cos\theta = 1\)
Using the identity \(\sin^2\theta + \cos^2\theta = 1\), we substitute 1 into the equation:
\(1 + 2\sin\theta\cos\theta = 1\)
Subtract 1 from both sides:
\(2\sin\theta\cos\theta = 0\)
Dividing by 2 gives:
\(\sin\theta\cos\theta = 0\)
Hence, proved.
In simple words: Square both sides of the given equation. Since \(\sin^2\theta + \cos^2\theta = 1\), you can cancel 1 from both sides, which leaves \(2\sin\theta\cos\theta = 0\), so the product is 0.

Exam Tip: Squaring both sides is a very common first step when given linear combinations of sine and cosine equal to a constant.

Question. Prove \(\frac{\sec A + \tan A - 1}{1 + \tan A - \sec A} = \frac{\cos A}{1 - \sin A}\) OR \(\frac{1 + \sin A}{\cos A}\) OR \(\sec A + \tan A\)
Answer:
Let us begin with the left hand side (LHS):
\(\text{LHS} = \frac{\sec A + \tan A - 1}{\tan A - \sec A + 1}\)
Using the identity \(1 = \sec^2 A - \tan^2 A\), we substitute this in place of the \(1\) in the numerator:
\(\text{LHS} = \frac{(\sec A + \tan A) - (\sec^2 A - \tan^2 A)}{\tan A - \sec A + 1}\)
Factorize \(\sec^2 A - \tan^2 A\) as \((\sec A - \tan A)(\sec A + \tan A)\):
\(\text{LHS} = \frac{(\sec A + \tan A) - (\sec A - \tan A)(\sec A + \tan A)}{\tan A - \sec A + 1}\)
Take \(\sec A + \tan A\) as a common factor in the numerator:
\(\text{LHS} = \frac{(\sec A + \tan A)[1 - (\sec A - \tan A)]}{\tan A - \sec A + 1}\)

\(\implies \text{LHS} = \frac{(\sec A + \tan A)(1 - \sec A + \tan A)}{1 + \tan A - \sec A}\)
Since \((1 - \sec A + \tan A)\) is identical to the denominator \((1 + \tan A - \sec A)\), they cancel out:
\(\text{LHS} = \sec A + \tan A\) (Third form proved)

Now, express this in terms of sine and cosine:
\(\sec A + \tan A = \frac{1}{\cos A} + \frac{\sin A}{\cos A} = \frac{1 + \sin A}{\cos A}\) (Second form proved)

Next, multiply the numerator and denominator of this result by \((1 - \sin A)\):
\(\frac{1 + \sin A}{\cos A} \times \frac{1 - \sin A}{1 - \sin A} = \frac{1 - \sin^2 A}{\cos A(1 - \sin A)}\)
Since \(1 - \sin^2 A = \cos^2 A\):
\(\frac{\cos^2 A}{\cos A(1 - \sin A)} = \frac{\cos A}{1 - \sin A}\) (First form proved)
Thus, all three forms are proved.
In simple words: Replace the 1 in the numerator with \(\sec^2 A - \tan^2 A\) and factor it. Cancel out the matching denominator to get \(\sec A + \tan A\), then rewrite it to find the other two equivalents.

Exam Tip: Substituting \(1 = \sec^2 A - \tan^2 A\) (or \(\csc^2 A - \cot^2 A\)) only in either the numerator or the denominator is a key trick for this type of problem.

Question. Prove \(\frac{1+\sin A + \cos A}{1 - \sin A + \cos A} = \sec A + \tan A\)
Answer:
Let us group the terms in the left hand side (LHS) as follows:
\(\text{LHS} = \frac{(1 + \cos A) + \sin A}{(1 + \cos A) - \sin A}\)
Multiply the numerator and the denominator by the conjugate of the denominator, \((1 + \cos A) + \sin A\):
\(\text{LHS} = \frac{[(1 + \cos A) + \sin A]^2}{[(1 + \cos A) - \sin A][(1 + \cos A) + \sin A]}\)
First, expand the numerator:
\([(1 + \cos A) + \sin A]^2 = (1 + \cos A)^2 + \sin^2 A + 2\sin A(1 + \cos A)\)
\(= 1 + \cos^2 A + 2\cos A + \sin^2 A + 2\sin A + 2\sin A\cos A\)
Substitute \(\sin^2 A + \cos^2 A = 1\):
\(= 1 + 1 + 2\cos A + 2\sin A + 2\sin A\cos A\)
\(= 2 + 2\cos A + 2\sin A + 2\sin A\cos A\)
Factor by grouping:
\(= 2(1 + \cos A) + 2\sin A(1 + \cos A) = 2(1 + \cos A)(1 + \sin A)\)

Next, expand the denominator:
\([(1 + \cos A) - \sin A][(1 + \cos A) + \sin A] = (1 + \cos A)^2 - \sin^2 A\)
\(= 1 + \cos^2 A + 2\cos A - (1 - \cos^2 A)\)
\(= 2\cos^2 A + 2\cos A = 2\cos A(1 + \cos A)\)

Now substitute both back into the fraction:
\(\text{LHS} = \frac{2(1 + \cos A)(1 + \sin A)}{2\cos A(1 + \cos A)}\)
Cancel the common terms \(2\) and \((1 + \cos A)\):
\(\text{LHS} = \frac{1 + \sin A}{\cos A} = \frac{1}{\cos A} + \frac{\sin A}{\cos A} = \sec A + \tan A = \text{RHS}\)
Hence, proved.
In simple words: Group \((1+\cos A)\) as one block and multiply the top and bottom by \((1+\cos A) + \sin A\). Simplify both parts and cancel the common factors to get \(\sec A + \tan A\).

Exam Tip: Grouping terms strategically (like combining \(1\) and \(\cos A\)) helps to apply binomial expansion and difference of squares efficiently.

Question. Prove \(\frac{\cot A + \csc A - 1}{\cot A - \csc A + 1} = \csc A + \cot A\)
Answer:
Let us write the left hand side (LHS) as:
\(\text{LHS} = \frac{\cot A + \csc A - 1}{\cot A - \csc A + 1}\)
We know the identity \(1 = \csc^2 A - \cot^2 A\). Let us substitute this for the \(1\) in the numerator:
\(\text{LHS} = \frac{(\csc A + \cot A) - (\csc^2 A - \cot^2 A)}{\cot A - \csc A + 1}\)
Factorize the second term in the numerator using the difference of squares:
\(\text{LHS} = \frac{(\csc A + \cot A) - (\csc A - \cot A)(\csc A + \cot A)}{\cot A - \csc A + 1}\)
Take \(( \csc A + \cot A )\) as a common factor in the numerator:
\(\text{LHS} = \frac{(\csc A + \cot A)[1 - (\csc A - \cot A)]}{\cot A - \csc A + 1}\)

\(\implies \text{LHS} = \frac{(\csc A + \cot A)(1 - \csc A + \cot A)}{\cot A - \csc A + 1}\)
Since the terms \((1 - \csc A + \cot A)\) and \(( \cot A - \csc A + 1 )\) are identical, they cancel out:
\(\text{LHS} = \csc A + \cot A = \text{RHS}\)
Hence, proved.
In simple words: Replace the 1 in the top part with \(\csc^2 A - \cot^2 A\), factor it out, and then cancel the common block from both the numerator and denominator.

Exam Tip: Be careful with the signs when expanding the parentheses after factoring out the common term in the numerator.

Question. Prove \(\frac{\cos A - \sin A + 1}{\cos A + \sin A - 1} = \csc A + \cot A\)
Answer:
Let us start with the left hand side (LHS):
\(\text{LHS} = \frac{\cos A - \sin A + 1}{\cos A + \sin A - 1}\)
To express this in terms of cosecant and cotangent, let us divide both the numerator and denominator by \(\sin A\):
\(\text{LHS} = \frac{\frac{\cos A}{\sin A} - \frac{\sin A}{\sin A} + \frac{1}{\sin A}}{\frac{\cos A}{\sin A} + \frac{\sin A}{\sin A} - \frac{1}{\sin A}}\)

\(\implies \text{LHS} = \frac{\cot A - 1 + \csc A}{\cot A + 1 - \csc A} = \frac{\cot A + \csc A - 1}{\cot A - \csc A + 1}\)
Using the identity \(1 = \csc^2 A - \cot^2 A\), we substitute it in the numerator:
\(\text{LHS} = \frac{(\cot A + \csc A) - (\csc^2 A - \cot^2 A)}{\cot A - \csc A + 1}\)
Factorize using difference of squares:
\(\text{LHS} = \frac{(\csc A + \cot A) - (\csc A - \cot A)(\csc A + \cot A)}{\cot A - \csc A + 1}\)
Take \(( \csc A + \cot A )\) common in the numerator:
\(\text{LHS} = \frac{(\csc A + \cot A)(1 - \csc A + \cot A)}{\cot A - \csc A + 1}\)
Cancel the common term in the numerator and denominator:
\(\text{LHS} = \csc A + \cot A = \text{RHS}\)
Hence, proved.
In simple words: Divide everything by \(\sin A\) to change the terms into \(\cot A\) and \(\csc A\). Then, replace the 1 in the numerator with \(\csc^2 A - \cot^2 A\) and simplify.

Exam Tip: If a proof has \(\csc A\) and \(\cot A\) on one side but \(\sin A\) and \(\cos A\) on the other, dividing the terms by \(\sin A\) is the most efficient starting step.

Question. Prove \(\frac{\sin A + \cos A}{\sin A - \cos A} + \frac{\sin A - \cos A}{\sin A + \cos A} = \frac{2\sec^2 A}{\tan^2 A - 1}\)
Answer:
Let us combine the two fractions on the left hand side (LHS) by taking a common denominator:
\(\text{LHS} = \frac{(\sin A + \cos A)^2 + (\sin A - \cos A)^2}{(\sin A - \cos A)(\sin A + \cos A)}\)
Expand the terms in the numerator:
\(\text{Numerator} = (\sin^2 A + \cos^2 A + 2\sin A\cos A) + (\sin^2 A + \cos^2 A - 2\sin A\cos A)\)
\(= 2(\sin^2 A + \cos^2 A)\)
Using the identity \(\sin^2 A + \cos^2 A = 1\):
\(\text{Numerator} = 2(1) = 2\)

Simplify the denominator:
\(\text{Denominator} = \sin^2 A - \cos^2 A\)
Thus, the expression becomes:
\(\text{LHS} = \frac{2}{\sin^2 A - \cos^2 A}\)
To get the expression in terms of secant and tangent, divide both the numerator and the denominator by \(\cos^2 A\):
\(\text{LHS} = \frac{\frac{2}{\cos^2 A}}{\frac{\sin^2 A}{\cos^2 A} - \frac{\cos^2 A}{\cos^2 A}}\)

\(\implies \text{LHS} = \frac{2\sec^2 A}{\tan^2 A - 1} = \text{RHS}\)
Hence, proved.
In simple words: Combine the fractions by cross-multiplying. Use standard expansion to simplify the top to 2, then divide both top and bottom by \(\cos^2 A\) to change the terms into secant and tangent.

Exam Tip: Expanding \((a+b)^2 + (a-b)^2\) always simplifies cleanly to \(2(a^2 + b^2)\), which is a useful algebraic shortcut to remember.

Question. Prove \(\frac{\tan A}{1+\cot A} + \frac{\cot A}{1+\tan A} = \sec A \csc A - 1\)
Answer:
Starting with the left hand side (LHS):
\(\text{LHS} = \frac{\tan A}{1+\cot A} + \frac{\cot A}{1+\tan A}\)
Convert \(\cot A\) to \(\frac{1}{\tan A}\):
\(\text{LHS} = \frac{\tan A}{1+\frac{1}{\tan A}} + \frac{\frac{1}{\tan A}}{1+\tan A}\)

\(\implies \text{LHS} = \frac{\tan^2 A}{\tan A + 1} + \frac{1}{\tan A(1+\tan A)}\)
Combine the fractions with a common denominator of \(\tan A(1+\tan A)\):
\(\text{LHS} = \frac{\tan^3 A + 1}{\tan A(1+\tan A)}\)
Use the algebraic identity \(x^3 + 1 = (x+1)(x^2 - x + 1)\) to factorize the numerator:
\(\text{LHS} = \frac{(\tan A + 1)(\tan^2 A - \tan A + 1)}{\tan A(1+\tan A)}\)
Cancel out the common term \((1+\tan A)\):
\(\text{LHS} = \frac{\tan^2 A - \tan A + 1}{\tan A}\)

\(\implies \text{LHS} = \frac{\tan^2 A}{\tan A} - \frac{\tan A}{\tan A} + \frac{1}{\tan A}\)

\(\implies \text{LHS} = \tan A - 1 + \cot A\)
Rearranging the terms:
\(\text{LHS} = (\tan A + \cot A) - 1\)
Now, let us simplify \(\tan A + \cot A\):
\(\tan A + \cot A = \frac{\sin A}{\cos A} + \frac{\cos A}{\sin A} = \frac{\sin^2 A + \cos^2 A}{\sin A \cos A} = \frac{1}{\sin A \cos A} = \sec A \csc A\)
Substitute this back into the expression:
\(\text{LHS} = \sec A \csc A - 1 = \text{RHS}\)
Hence, proved.
In simple words: Write \(\cot A\) as \(1/\tan A\). Bring the terms to a common denominator, factor the numerator using the sum of cubes, cancel the common factor, and then rewrite \(\tan A + \cot A\) as \(\sec A \csc A\).

Exam Tip: Whenever you see both \(\tan\) and \(\cot\) in a fraction, expressing everything in terms of \(\tan\) first can drastically simplify the algebra compared to converting to \(\sin\) and \(\cos\) immediately.

Question. Prove \((\frac{1}{\cos\theta} - \cos\theta)(\frac{1}{\sin\theta} - \sin\theta) = \frac{1}{\tan\theta + \cot\theta}\)
Answer:
Let us simplify the left hand side (LHS) first:
\(\text{LHS} = \left(\frac{1}{\cos\theta} - \cos\theta\right)\left(\frac{1}{\sin\theta} - \sin\theta\right)\)
Take the common denominator inside each parenthesis:
\(\text{LHS} = \left(\frac{1 - \cos^2\theta}{\cos\theta}\right)\left(\frac{1 - \sin^2\theta}{\sin\theta}\right)\)
Apply the identities \(1 - \cos^2\theta = \sin^2\theta\) and \(1 - \sin^2\theta = \cos^2\theta\):
\(\text{LHS} = \left(\frac{\sin^2\theta}{\cos\theta}\right)\left(\frac{\cos^2\theta}{\sin\theta}\right)\)

\(\implies \text{LHS} = \sin\theta \cos\theta\)

Now, let us simplify the right hand side (RHS):
\(\text{RHS} = \frac{1}{\tan\theta + \cot\theta}\)
Convert the denominator to sine and cosine:
\(\text{RHS} = \frac{1}{\frac{\sin\theta}{\cos\theta} + \frac{\cos\theta}{\sin\theta}}\)

\(\implies \text{RHS} = \frac{1}{\frac{\sin^2\theta + \cos^2\theta}{\sin\theta\cos\theta}}\)
Substitute \(\sin^2\theta + \cos^2\theta = 1\):
\(\text{RHS} = \frac{1}{\frac{1}{\sin\theta\cos\theta}} = \sin\theta\cos\theta\)
Since \(\text{LHS} = \text{RHS}\), the identity is verified.
In simple words: Simplify both sides. The left side becomes \(\sin\theta\cos\theta\) after using basic identity conversions, and the right side also simplifies to \(\sin\theta\cos\theta\) after writing \(\tan\theta\) and \(\cot\theta\) in terms of sine and cosine.

Exam Tip: Don't hesitate to solve the left-hand side and right-hand side separately when a direct transformation from one to the other is not obvious.

Question. Prove \(\frac{\tan^2\phi}{\tan^2\phi - 1} + \frac{\cos^2\phi}{\sin^2\phi - \cos^2\phi} = \frac{1}{\sin^2\phi - \cos^2\phi}\)
Answer:
Let us simplify the first term of the left hand side (LHS):
\(\frac{\tan^2\phi}{\tan^2\phi - 1} = \frac{\frac{\sin^2\phi}{\cos^2\phi}}{\frac{\sin^2\phi}{\cos^2\phi} - 1}\)
Take the common denominator in the denominator of this fraction:
\(= \frac{\frac{\sin^2\phi}{\cos^2\phi}}{\frac{\sin^2\phi - \cos^2\phi}{\cos^2\phi}}\)
Cancel out the \(\cos^2\phi\) terms from the denominators:
\(= \frac{\sin^2\phi}{\sin^2\phi - \cos^2\phi}\)

Now, substitute this back into the LHS:
\(\text{LHS} = \frac{\sin^2\phi}{\sin^2\phi - \cos^2\phi} + \frac{\cos^2\phi}{\sin^2\phi - \cos^2\phi}\)
Combine the fractions under the common denominator \(\sin^2\phi - \cos^2\phi\):
\(\text{LHS} = \frac{\sin^2\phi + \cos^2\phi}{\sin^2\phi - \cos^2\phi}\)
Since \(\sin^2\phi + \cos^2\phi = 1\), we get:
\(\text{LHS} = \frac{1}{\sin^2\phi - \cos^2\phi} = \text{RHS}\)
Hence, proved.
In simple words: Convert the tangent term in the first fraction into sine and cosine. This gives both fractions the exact same denominator, making it easy to combine them and simplify the top to 1.

Exam Tip: When denominators look similar but have different trigonometric functions, converting to sines and cosines often yields a common denominator instantly.

Question. Prove \((\sin\theta + \csc\theta)^2 + (\cos\theta + \sec\theta)^2 = \tan^2\theta + \cot^2\theta + 7\)
Answer:
Let us expand the terms on the left hand side (LHS) using the algebraic identity \((a+b)^2 = a^2 + 2ab + b^2\):
\(\text{LHS} = (\sin^2\theta + \csc^2\theta + 2\sin\theta\csc\theta) + (\cos^2\theta + \sec^2\theta + 2\cos\theta\sec\theta)\)
We know the reciprocal relationships:
\(\sin\theta\csc\theta = 1\) and \(\cos\theta\sec\theta = 1\)
Substituting these values back in:
\(\text{LHS} = \sin^2\theta + \csc^2\theta + 2(1) + \cos^2\theta + \sec^2\theta + 2(1)\)

\(\implies \text{LHS} = \sin^2\theta + \cos^2\theta + \csc^2\theta + \sec^2\theta + 4\)
Using the identity \(\sin^2\theta + \cos^2\theta = 1\):
\(\text{LHS} = 1 + \csc^2\theta + \sec^2\theta + 4 = 5 + \csc^2\theta + \sec^2\theta\)
Now, use the identities \(\csc^2\theta = 1 + \cot^2\theta\) and \(\sec^2\theta = 1 + \tan^2\theta\):
\(\text{LHS} = 5 + (1 + \cot^2\theta) + (1 + \tan^2\theta)\)

\(\implies \text{LHS} = \tan^2\theta + \cot^2\theta + 7 = \text{RHS}\)
Hence, proved.
In simple words: Square both brackets using the binomial formula. Use the reciprocal identities to simplify the cross-product terms to integers, combine the remaining squares, and convert cosecant and secant into cotangent and tangent.

Exam Tip: Recognizing standard products like \(\sin\theta\csc\theta = 1\) early prevents unnecessary conversion to fractions and keeps the proof clean.

Question. Prove \(\frac{\sin^2 A}{\cos^2 A} + \frac{\cos^2 A}{\sin^2 A} = \frac{1}{\sin^2 A\cos^2 A} - 2\)
Answer:
Let us combine the terms on the left hand side (LHS) by finding a common denominator:
\(\text{LHS} = \frac{\sin^4 A + \cos^4 A}{\sin^2 A \cos^2 A}\)
We can rewrite the numerator using the algebraic identity \(a^2 + b^2 = (a+b)^2 - 2ab\), by letting \(a = \sin^2 A\) and \(b = \cos^2 A\):
\(\sin^4 A + \cos^4 A = (\sin^2 A + \cos^2 A)^2 - 2\sin^2 A \cos^2 A\)
Since \(\sin^2 A + \cos^2 A = 1\), this simplifies to:
\(\sin^4 A + \cos^4 A = 1^2 - 2\sin^2 A \cos^2 A = 1 - 2\sin^2 A \cos^2 A\)
Now, substitute this back into our LHS expression:
\(\text{LHS} = \frac{1 - 2\sin^2 A \cos^2 A}{\sin^2 A \cos^2 A}\)
Split this into two separate fractions:
\(\text{LHS} = \frac{1}{\sin^2 A \cos^2 A} - \frac{2\sin^2 A \cos^2 A}{\sin^2 A \cos^2 A}\)

\(\implies \text{LHS} = \frac{1}{\sin^2 A \cos^2 A} - 2 = \text{RHS}\)
Hence, proved.
In simple words: Combine the two fractions on the left. Express \(\sin^4 A + \cos^4 A\) as \((1 - 2\sin^2 A\cos^2 A)\) using algebraic formulas, and split the fraction to get the right side.

Exam Tip: The identity \(\sin^4 \theta + \cos^4 \theta = 1 - 2\sin^2 \theta \cos^2 \theta\) is extremely common in high-scoring trigonometry questions; memorize it to save time.

Question. Prove \((\sec A + \cos A)(\sec A - \cos A) = \tan^2 A + \sin^2 A\)
Answer:
Starting with the left hand side (LHS):
\(\text{LHS} = (\sec A + \cos A)(\sec A - \cos A)\)
Using the difference of squares identity \((a+b)(a-b) = a^2 - b^2\):
\(\text{LHS} = \sec^2 A - \cos^2 A\)
Now, we convert \(\sec^2 A\) and \(\cos^2 A\) using fundamental identities:
\(\sec^2 A = 1 + \tan^2 A\)
\(\cos^2 A = 1 - \sin^2 A\)
Substitute these back into the expression:
\(\text{LHS} = (1 + \tan^2 A) - (1 - \sin^2 A)\)

\(\implies \text{LHS} = 1 + \tan^2 A - 1 + \sin^2 A\)

\(\implies \text{LHS} = \tan^2 A + \sin^2 A = \text{RHS}\)
Hence, proved.
In simple words: Multiply the brackets using the difference of squares rule. Change secant squared to \(1 + \tan^2 A\) and cosine squared to \(1 - \sin^2 A\) so that the 1s cancel out.

Exam Tip: Be mindful of parentheses when subtracting \((1 - \sin^2 A)\) to ensure the sign changes correctly to plus.

Question. Prove \(\sec A(1 - \sin A)(\sec A + \tan A) = 1\)
Answer:
Let us simplify the left hand side (LHS):
\(\text{LHS} = \sec A(1 - \sin A)(\sec A + \tan A)\)
Distribute the first term \(\sec A\) inside the first parentheses:
\(\text{LHS} = (\sec A - \sec A\sin A)(\sec A + \tan A)\)
Since \(\sec A = \frac{1}{\cos A}\), we have \(\sec A \sin A = \frac{\sin A}{\cos A} = \tan A\):
\(\text{LHS} = (\sec A - \tan A)(\sec A + \tan A)\)
Use the difference of squares identity \((a-b)(a+b) = a^2 - b^2\):
\(\text{LHS} = \sec^2 A - \tan^2 A\)
Since we know \(\sec^2 A - \tan^2 A = 1\):
\(\text{LHS} = 1 = \text{RHS}\)
Hence, proved.
In simple words: Multiply \(\sec A\) into the first bracket to get \(\sec A - \tan A\). Then multiply this with the second bracket, which gives \(\sec^2 A - \tan^2 A\), which is equal to 1.

Exam Tip: Expanding term-by-term selectively rather than changing everything to sines and cosines right away can often lead to a much quicker solution.

Question. Prove \(\frac{1 + \tan^2 A}{1 + \cot^2 A} = \left(\frac{1 - \tan A}{1 - \cot A}\right)^2 = \tan^2 A\)
Answer:
Let us prove that the first expression equals \(\tan^2 A\):
\(\text{First Expression} = \frac{1 + \tan^2 A}{1 + \cot^2 A}\)
Using the identities \(1 + \tan^2 A = \sec^2 A\) and \(1 + \cot^2 A = \csc^2 A\):
\(\text{First Expression} = \frac{\sec^2 A}{\csc^2 A}\)
Convert to sines and cosines:
\(\text{First Expression} = \frac{\frac{1}{\cos^2 A}}{\frac{1}{\sin^2 A}} = \frac{\sin^2 A}{\cos^2 A} = \tan^2 A\)

Now, let us prove that the second expression equals \(\tan^2 A\):
\(\text{Second Expression} = \left(\frac{1 - \tan A}{1 - \cot A}\right)^2\)
Convert \(\cot A\) to \(\frac{1}{\tan A}\):
\(\text{Second Expression} = \left(\frac{1 - \tan A}{1 - \frac{1}{\tan A}}\right)^2\)
Simplify the denominator inside the parentheses:
\(\text{Second Expression} = \left(\frac{1 - \tan A}{\frac{\tan A - 1}{\tan A}}\right)^2\)
Note that \(\tan A - 1 = -(1 - \tan A)\):
\(\text{Second Expression} = \left(\frac{1 - \tan A}{-\frac{1 - \tan A}{\tan A}}\right)^2 = \left(-\tan A\right)^2 = \tan^2 A\)
Since both expressions simplify to \(\tan^2 A\), the identity is fully proved.
In simple words: First, use \(\sec^2 A\) and \(\csc^2 A\) to simplify the left fraction to \(\tan^2 A\). Then write \(\cot A\) as \(1/\tan A\) in the middle bracket, cancel out terms, and square the remaining \(-\tan A\) to get \(\tan^2 A\).

Exam Tip: Remember that squaring a negative term like \((-\tan A)^2\) yields a positive result, \(\tan^2 A\).

Question. Prove \(\frac{\sec\theta + \tan\theta}{\sec\theta - \tan\theta} = \left(\frac{1 + \sin\theta}{\cos\theta}\right)^2\)
Answer:
Let us work with the left hand side (LHS):
\(\text{LHS} = \frac{\sec\theta + \tan\theta}{\sec\theta - \tan\theta}\)
Convert secant and tangent to sine and cosine:
\(\text{LHS} = \frac{\frac{1}{\cos\theta} + \frac{\sin\theta}{\cos\theta}}{\frac{1}{\cos\theta} - \frac{\sin\theta}{\cos\theta}} = \frac{\frac{1+\sin\theta}{\cos\theta}}{\frac{1-\sin\theta}{\cos\theta}}\)

\(\implies \text{LHS} = \frac{1+\sin\theta}{1-\sin\theta}\)
Multiply both the numerator and the denominator by \((1+\sin\theta)\):
\(\text{LHS} = \frac{(1+\sin\theta)^2}{(1-\sin\theta)(1+\sin\theta)}\)

\(\implies \text{LHS} = \frac{(1+\sin\theta)^2}{1-\sin^2\theta}\)
Using the identity \(1-\sin^2\theta = \cos^2\theta\):
\(\text{LHS} = \frac{(1+\sin\theta)^2}{\cos^2\theta} = \left(\frac{1+\sin\theta}{\cos\theta}\right)^2 = \text{RHS}\)
Hence, proved.
In simple words: Convert the left side to sines and cosines to get \(\frac{1+\sin\theta}{1-\sin\theta}\). Then, multiply the top and bottom by \((1+\sin\theta)\) to form \(\cos^2\theta\) on the bottom, giving you the squared bracket on the right.

Exam Tip: Converting to sines and cosines first usually makes it obvious how to rationalize the denominator or numerator.

Question. Prove \(\frac{\tan\theta + \sin\theta}{\tan\theta - \sin\theta} = \frac{\sec\theta + 1}{\sec\theta - 1}\)
Answer:
Starting with the left hand side (LHS):
\(\text{LHS} = \frac{\tan\theta + \sin\theta}{\tan\theta - \sin\theta}\)
Substitute \(\tan\theta = \frac{\sin\theta}{\cos\theta}\) into the expression:
\(\text{LHS} = \frac{\frac{\sin\theta}{\cos\theta} + \sin\theta}{\frac{\sin\theta}{\cos\theta} - \sin\theta}\)
Factor out \(\sin\theta\) from both the numerator and denominator:
\(\text{LHS} = \frac{\sin\theta \left(\frac{1}{\cos\theta} + 1\right)}{\sin\theta \left(\frac{1}{\cos\theta} - 1\right)}\)
Cancel out the common \(\sin\theta\) factor:
\(\text{LHS} = \frac{\frac{1}{\cos\theta} + 1}{\frac{1}{\cos\theta} - 1}\)
Since \(\frac{1}{\cos\theta} = \sec\theta\), we can substitute it back:
\(\text{LHS} = \frac{\sec\theta + 1}{\sec\theta - 1} = \text{RHS}\)
Hence, proved.
In simple words: Rewrite \(\tan\theta\) as \(\sin\theta/\cos\theta\). Factor out \(\sin\theta\) from both the top and bottom to cancel them, which leaves you with terms that convert directly to \(\sec\theta\).

Exam Tip: Factoring out common trigonometric functions (like \(\sin\theta\) here) is an excellent way to simplify fractions quickly.

Question. Prove \(\frac{\sin\theta}{1 - \cot\theta} + \frac{\cos\theta}{1 - \tan\theta} = \sin\theta + \cos\theta\)
Answer:
Let us begin with the left hand side (LHS):
\(\text{LHS} = \frac{\sin\theta}{1 - \cot\theta} + \frac{\cos\theta}{1 - \tan\theta}\)
Convert \(\cot\theta\) and \(\tan\theta\) into sine and cosine:
\(\text{LHS} = \frac{\sin\theta}{1 - \frac{\cos\theta}{\sin\theta}} + \frac{\cos\theta}{1 - \frac{\sin\theta}{\cos\theta}}\)
Simplify the denominators of both fractions:
\(\text{LHS} = \frac{\sin^2\theta}{\sin\theta - \cos\theta} + \frac{\cos^2\theta}{\cos\theta - \sin\theta}\)
Rewrite the second denominator as \(-(\sin\theta - \cos\theta)\) to get a common denominator:
\(\text{LHS} = \frac{\sin^2\theta}{\sin\theta - \cos\theta} - \frac{\cos^2\theta}{\sin\theta - \cos\theta}\)

\(\implies \text{LHS} = \frac{\sin^2\theta - \cos^2\theta}{\sin\theta - \cos\theta}\)
Factor the numerator using the difference of squares identity \(a^2 - b^2 = (a-b)(a+b)\):
\(\text{LHS} = \frac{(\sin\theta - \cos\theta)(\sin\theta + \cos\theta)}{\sin\theta - \cos\theta}\)
Cancel out the common term \(( \sin\theta - \cos\theta )\):
\(\text{LHS} = \sin\theta + \cos\theta = \text{RHS}\)
Hence, proved.
In simple words: Convert everything to sines and cosines. Make the denominators of the two fractions match by changing the sign of one, then combine them and factor the top to cancel out the denominator.

Exam Tip: Watch out for denominators like \((x-y)\) and \((y-x)\); you can make them identical simply by pulling out a minus sign.

Question. If \(\frac{\cos\alpha}{\cos\beta} = m ; \frac{\cos\alpha}{\sin\beta} = n\) Show that \((m^2 + n^2)\cos^2\beta = n^2\)
Answer:
We are given:
\(m = \frac{\cos\alpha}{\cos\beta}\)
\(n = \frac{\cos\alpha}{\sin\beta}\)
Let us substitute these values of \(m\) and \(n\) into the left hand side (LHS) of the equation:
\(\text{LHS} = (m^2 + n^2)\cos^2\beta\)

\(\implies \text{LHS} = \left(\frac{\cos^2\alpha}{\cos^2\beta} + \frac{\cos^2\alpha}{\sin^2\beta}\right)\cos^2\beta\)
Factor out \(\cos^2\alpha\) from the terms inside the parentheses:
\(\text{LHS} = \cos^2\alpha \left(\frac{1}{\cos^2\beta} + \frac{1}{\sin^2\beta}\right)\cos^2\beta\)
Find a common denominator for the terms inside the bracket:
\(\text{LHS} = \cos^2\alpha \left(\frac{\sin^2\beta + \cos^2\beta}{\cos^2\beta \sin^2\beta}\right)\cos^2\beta\)
Using the identity \(\sin^2\beta + \cos^2\beta = 1\):
\(\text{LHS} = \cos^2\alpha \left(\frac{1}{\cos^2\beta \sin^2\beta}\right)\cos^2\beta\)
Cancel out \(\cos^2\beta\) from the numerator and denominator:
\(\text{LHS} = \frac{\cos^2\alpha}{\sin^2\beta}\)
This can be written as:
\(\text{LHS} = \left(\frac{\cos\alpha}{\sin\beta}\right)^2 = n^2 = \text{RHS}\)
Hence, shown.
In simple words: Substitute the given expressions for \(m\) and \(n\) into the formula. Group the terms, simplify using the identity \(\sin^2\beta + \cos^2\beta = 1\), cancel the common terms, and you will get \(n^2\).

Exam Tip: Factoring out terms like \(\cos^2\alpha\) early in algebraic substitutions makes handling fractions much cleaner and reduces mistakes.

Question. If \(x = a\cos\theta - b\sin\theta\)
\(y = a\sin\theta + b\cos\theta\)
Prove \(x^2 + y^2 = a^2 + b^2\)
Answer:
Let us calculate the squares of \(x\) and \(y\):
\(x^2 = (a\cos\theta - b\sin\theta)^2 = a^2\cos^2\theta + b^2\sin^2\theta - 2ab\sin\theta\cos\theta\)
\(y^2 = (a\sin\theta + b\cos\theta)^2 = a^2\sin^2\theta + b^2\cos^2\theta + 2ab\sin\theta\cos\theta\)
Now, let us add these two squared expressions together:
\(x^2 + y^2 = (a^2\cos^2\theta + b^2\sin^2\theta - 2ab\sin\theta\cos\theta) + (a^2\sin^2\theta + b^2\cos^2\theta + 2ab\sin\theta\cos\theta)\)
Notice that the cross-multiplied terms \(-2ab\sin\theta\cos\theta\) and \(+2ab\sin\theta\cos\theta\) cancel each other out:
\(x^2 + y^2 = a^2\cos^2\theta + b^2\sin^2\theta + a^2\sin^2\theta + b^2\cos^2\theta\)
Group the terms containing \(a^2\) and \(b^2\) separately:
\(x^2 + y^2 = a^2(\cos^2\theta + \sin^2\theta) + b^2(\sin^2\theta + \cos^2\theta)\)
Using the identity \(\sin^2\theta + \cos^2\theta = 1\):
\(x^2 + y^2 = a^2(1) + b^2(1) = a^2 + b^2\)
Hence, proved.
In simple words: Square both equations. When you add them, the mixed middle terms cancel out. Group \(a^2\) and \(b^2\) to simplify using \(\sin^2\theta + \cos^2\theta = 1\).

Exam Tip: In algebraic trigonometry problems, expanding squares and summing them up is a very standard method to eliminate trigonometric terms.

Question. If \(x\sin^3\theta + y\cos^3\theta = \sin\theta\cos\theta\)
And \(x\sin\theta = y\cos\theta\)
Show \(x^2 + y^2 = 1\)
Answer:
We are given the following two equations:
\(x\sin^3\theta + y\cos^3\theta = \sin\theta\cos\theta\) - (Equation 1)
\(x\sin\theta = y\cos\theta\) - (Equation 2)

Let us rewrite Equation 1 by splitting the cubed terms:
\((x\sin\theta)\sin^2\theta + (y\cos\theta)\cos^2\theta = \sin\theta\cos\theta\)
Substitute \(y\cos\theta = x\sin\theta\) from Equation 2 into the second term:
\((x\sin\theta)\sin^2\theta + (x\sin\theta)\cos^2\theta = \sin\theta\cos\theta\)
Factor out \(x\sin\theta\) from the left hand side:
\(x\sin\theta (\sin^2\theta + \cos^2\theta) = \sin\theta\cos\theta\)
Using the identity \(\sin^2\theta + \cos^2\theta = 1\):
\(x\sin\theta (1) = \sin\theta\cos\theta\)

\(\implies x\sin\theta = \sin\theta\cos\theta\)
Dividing both sides by \(\sin\theta\) yields:
\(x = \cos\theta\)

Now, substitute \(x = \cos\theta\) back into Equation 2:
\((\cos\theta)\sin\theta = y\cos\theta\)
Dividing both sides by \(\cos\theta\) yields:
\(y = \sin\theta\)

Finally, calculate \(x^2 + y^2\):
\(x^2 + y^2 = \cos^2\theta + \sin^2\theta = 1\)
Hence, shown.
In simple words: Break down the cubed terms in the first equation. Use the second equation to replace \(y\cos\theta\) with \(x\sin\theta\), which simplifies the equation to \(x = \cos\theta\). Then find that \(y = \sin\theta\) and sum their squares to get 1.

Exam Tip: Substituting a simpler relation into a higher-degree trigonometric equation is a highly effective way to eliminate variables.