CBSE Class 10 Mathematics Trigonometry Assignment Set 10

Prove the following:

Question. \( \sin 20^\circ \sin 70^\circ - \cos 20^\circ \cos 70^\circ = 0 \)
Answer: We want to show that \( \sin 20^\circ \sin 70^\circ - \cos 20^\circ \cos 70^\circ = 0 \). Let's express the complementary angles in terms of a single angle, say \( 20^\circ \). Using complementary angle relationships, we know: \( \sin 70^\circ = \sin(90^\circ - 20^\circ) = \cos 20^\circ \) \( \cos 70^\circ = \cos(90^\circ - 20^\circ) = \sin 20^\circ \) Now, substitute these terms into the Left-Hand Side (L.H.S.) of the expression: \( \text{L.H.S.} = \sin 20^\circ \cos 20^\circ - \cos 20^\circ \sin 20^\circ \)
\( \implies \text{L.H.S.} = \sin 20^\circ \cos 20^\circ - \sin 20^\circ \cos 20^\circ = 0 \) Since L.H.S. is equal to the Right-Hand Side (R.H.S.), the identity is proven.
In simple words: Swap the \( 70^\circ \) angles to \( 20^\circ \) using complementary rules. Subtracting the two identical terms results in zero.

Exam Tip: When proving trigonometric equations, always write out the formula you use to change the angles, as examiners look for these steps to award full marks.

Question. \( \cos 39^\circ \cos 51^\circ - \sin 39^\circ \sin 51^\circ = 0 \)
Answer: We want to prove that \( \cos 39^\circ \cos 51^\circ - \sin 39^\circ \sin 51^\circ = 0 \). Let's change the terms containing \( 51^\circ \) to terms with \( 39^\circ \) using complementary angle formulas: \( \cos 51^\circ = \cos(90^\circ - 39^\circ) = \sin 39^\circ \) \( \sin 51^\circ = \sin(90^\circ - 39^\circ) = \cos 39^\circ \) Now, substitute these back into the Left-Hand Side (L.H.S.): \( \text{L.H.S.} = \cos 39^\circ \sin 39^\circ - \sin 39^\circ \cos 39^\circ \)
\( \implies \text{L.H.S.} = \sin 39^\circ \cos 39^\circ - \sin 39^\circ \cos 39^\circ = 0 \) This is equal to the Right-Hand Side (R.H.S.), so the identity is verified.
In simple words: Change the \( 51^\circ \) terms into \( 39^\circ \) terms using complementary identities. The two parts of the subtraction become identical, giving a final result of zero.

Exam Tip: Keep the angle that you decide to convert consistent throughout the equation to keep your steps straightforward and easy to follow.

Question. \( \frac{\sin 40^\circ}{\cos 50^\circ} \times \frac{\tan 44^\circ}{\cot 46^\circ} \times \frac{\sec 55^\circ}{\text{cosec } 35^\circ} = 1 \)
Answer: Let's simplify the Left-Hand Side (L.H.S.) expression: \( \text{L.H.S.} = \frac{\sin 40^\circ}{\cos 50^\circ} \times \frac{\tan 44^\circ}{\cot 46^\circ} \times \frac{\sec 55^\circ}{\text{cosec } 35^\circ} \) Let's convert the angles in the denominators to match the numerators: \( \cos 50^\circ = \cos(90^\circ - 40^\circ) = \sin 40^\circ \) \( \cot 46^\circ = \cot(90^\circ - 44^\circ) = \tan 44^\circ \) \( \text{cosec } 35^\circ = \text{cosec}(90^\circ - 55^\circ) = \sec 55^\circ \) Now, substitute these back into the original expression: \( \text{L.H.S.} = \frac{\sin 40^\circ}{\sin 40^\circ} \times \frac{\tan 44^\circ}{\tan 44^\circ} \times \frac{\sec 55^\circ}{\sec 55^\circ} \)
\( \implies \text{L.H.S.} = 1 \times 1 \times 1 = 1 \) Thus, L.H.S. = R.H.S., proving the identity.
In simple words: Change the denominator angles to match the numerator angles using complementary identities. This makes each fraction equal to 1, and multiplying them together gives 1.

Exam Tip: In fraction problems where the sum of numerator and denominator angles is \( 90^\circ \), converting just one of them will always help you cancel them out to 1.

Question. \( \cos^2 85^\circ - \sin^2 5^\circ = 0 \)
Answer: We want to verify that \( \cos^2 85^\circ - \sin^2 5^\circ = 0 \). Using complementary angle rules, we convert the first term's angle: \( \cos 85^\circ = \cos(90^\circ - 5^\circ) = \sin 5^\circ \) Squaring both sides of the equation gives: \( \cos^2 85^\circ = \sin^2 5^\circ \) Substituting this into the Left-Hand Side (L.H.S.): \( \text{L.H.S.} = \sin^2 5^\circ - \sin^2 5^\circ = 0 \) Since this matches the R.H.S., the proof is complete.
In simple words: Convert the cosine of \( 85^\circ \) into the sine of \( 5^\circ \). Squaring it and subtracting the other sine squared leaves you with nothing.

Exam Tip: Be sure to apply squaring properly after using the complementary angle identity, as \( \cos^2 \theta \) is simply \( (\cos \theta)^2 \).

Question. \( \frac{\sin \theta}{\cos(90^\circ - \theta)} + \frac{\cos \theta}{\sin(90^\circ - \theta)} + \frac{\tan \theta}{\cot(90^\circ - \theta)} = 3 \)
Answer: Let's evaluate the Left-Hand Side (L.H.S.) of the given expression: \( \text{L.H.S.} = \frac{\sin \theta}{\cos(90^\circ - \theta)} + \frac{\cos \theta}{\sin(90^\circ - \theta)} + \frac{\tan \theta}{\cot(90^\circ - \theta)} \) Using complementary angle formulas on the denominators: \( \cos(90^\circ - \theta) = \sin \theta \) \( \sin(90^\circ - \theta) = \cos \theta \) \( \cot(90^\circ - \theta) = \tan \theta \) Substitute these values into the expression: \( \text{L.H.S.} = \frac{\sin \theta}{\sin \theta} + \frac{\cos \theta}{\cos \theta} + \frac{\tan \theta}{\tan \theta} \)
\( \implies \text{L.H.S.} = 1 + 1 + 1 = 3 \) This equals the R.H.S., which completes our proof.
In simple words: Change the bottom of each fraction using complementary formulas. Each fraction simplifies to 1, and adding them together gives 3.

Exam Tip: Clear bracket simplification is key. Always replace terms like \( \cos(90^\circ - \theta) \) with \( \sin \theta \) immediately to make the expression look less complex.

Question. \( \sin \theta \cdot \cos(90^\circ - \theta) + \cos \theta \cdot \sin(90^\circ - \theta) = 1 \)
Answer: We need to show that \( \sin \theta \cdot \cos(90^\circ - \theta) + \cos \theta \cdot \sin(90^\circ - \theta) = 1 \). We can rewrite the complementary angles inside the expression: \( \cos(90^\circ - \theta) = \sin \theta \) \( \sin(90^\circ - \theta) = \cos \theta \) Substituting these into the Left-Hand Side (L.H.S.): \( \text{L.H.S.} = \sin \theta \cdot \sin \theta + \cos \theta \cdot \cos \theta \)
\( \implies \text{L.H.S.} = \sin^2 \theta + \cos^2 \theta \) Using the standard Pythagorean trigonometric identity, we have: \( \sin^2 \theta + \cos^2 \theta = 1 \)
\( \implies \text{L.H.S.} = 1 \) L.H.S. is equal to R.H.S., proving the identity.
In simple words: Replace the complementary terms to get sine squared plus cosine squared, which is a standard trigonometric identity that always equals 1.

Exam Tip: Always state the fundamental Pythagorean identity \( \sin^2 \theta + \cos^2 \theta = 1 \) explicitly on the side to justify your final step.

Question. \( \frac{\sin(90^\circ - A) \cos(90^\circ - A)}{\tan A} = \cos^2 A \)
Answer: Let's evaluate the Left-Hand Side (L.H.S.): \( \text{L.H.S.} = \frac{\sin(90^\circ - A) \cos(90^\circ - A)}{\tan A} \) Substitute the complementary angle formulas for the numerator terms: \( \sin(90^\circ - A) = \cos A \) \( \cos(90^\circ - A) = \sin A \) Now substitute these into the numerator: \( \text{L.H.S.} = \frac{\cos A \cdot \sin A}{\tan A} \) We know that \( \tan A = \frac{\sin A}{\cos A} \). Let's replace this in the denominator: \( \text{L.H.S.} = \frac{\cos A \cdot \sin A}{\frac{\sin A}{\cos A}} \)
\( \implies \text{L.H.S.} = \cos A \cdot \sin A \times \frac{\cos A}{\sin A} \) Cancelling \( \sin A \) from both numerator and denominator: \( \text{L.H.S.} = \cos A \cdot \cos A = \cos^2 A \) This matches the R.H.S. exactly, hence proved.
In simple words: Convert the top terms using complementary formulas and write tangent as sine over cosine. When you divide, the sine terms cancel out, leaving cosine squared.

Exam Tip: Converting tangent to \( \frac{\sin}{\cos} \) is a very useful technique when other terms in the expression are sines and cosines.

Question. \( \frac{\sin^2 69^\circ + \cos^2 21^\circ}{\sin^2 69^\circ + \sin^2 21^\circ} = 2 \cos^2 21^\circ \) or \( 2 \sin^2 69^\circ \)
Answer: Let's evaluate the Left-Hand Side (L.H.S.) of the expression: \( \text{L.H.S.} = \frac{\sin^2 69^\circ + \cos^2 21^\circ}{\sin^2 69^\circ + \sin^2 21^\circ} \) First, let us simplify the denominator using the complementary identity: \( \sin 21^\circ = \sin(90^\circ - 69^\circ) = \cos 69^\circ \) Thus, the denominator becomes: \( \sin^2 69^\circ + \sin^2 21^\circ = \sin^2 69^\circ + \cos^2 69^\circ = 1 \) Now, substitute this value back into the expression: \( \text{L.H.S.} = \sin^2 69^\circ + \cos^2 21^\circ \) We can now prove the two alternative required forms:
First Form: Prove \( \text{L.H.S.} = 2 \cos^2 21^\circ \) Convert \( \sin 69^\circ \) to its complement: \( \sin 69^\circ = \sin(90^\circ - 21^\circ) = \cos 21^\circ \) Substitute this in: \( \text{L.H.S.} = (\cos 21^\circ)^2 + \cos^2 21^\circ = 2 \cos^2 21^\circ \)
Second Form: Prove \( \text{L.H.S.} = 2 \sin^2 69^\circ \) Convert \( \cos 21^\circ \) to its complement: \( \cos 21^\circ = \cos(90^\circ - 69^\circ) = \sin 69^\circ \) Substitute this in: \( \text{L.H.S.} = \sin^2 69^\circ + (\sin 69^\circ)^2 = 2 \sin^2 69^\circ \) Both target expressions are successfully proven.
In simple words: First show the bottom equals 1. Then, by swapping either the sine term or the cosine term with its complement, you get two identical squared terms that add up together.

Exam Tip: If an identity asks you to prove two different forms, show both steps explicitly to guarantee you get complete marks on the exam.

Question. \( \frac{\sin(30^\circ - \theta) \cos(60^\circ - \theta)}{\sin(30^\circ + \theta) \cos(60^\circ + \theta)} = 1 \)
Answer: Let's simplify the Left-Hand Side (L.H.S.): \( \text{L.H.S.} = \frac{\sin(30^\circ - \theta) \cos(60^\circ - \theta)}{\sin(30^\circ + \theta) \cos(60^\circ + \theta)} \) Let's convert the numerator's factors to complementary angles: \( \sin(30^\circ - \theta) = \cos[90^\circ - (30^\circ - \theta)] = \cos(60^\circ + \theta) \) \( \cos(60^\circ - \theta) = \sin[90^\circ - (60^\circ - \theta)] = \sin(30^\circ + \theta) \) Substitute these back into the numerator: \( \text{L.H.S.} = \frac{\cos(60^\circ + \theta) \sin(30^\circ + \theta)}{\sin(30^\circ + \theta) \cos(60^\circ + \theta)} \) Since the numerator and denominator have the exact same terms, they cancel out: \( \text{L.H.S.} = \frac{\sin(30^\circ + \theta) \cos(60^\circ + \theta)}{\sin(30^\circ + \theta) \cos(60^\circ + \theta)} = 1 \) L.H.S. is equal to R.H.S., proving the statement.
In simple words: Turn the top terms into complementary forms. This makes the top and bottom of the fraction completely identical, so they divide to give 1.

Exam Tip: When dealing with variables inside brackets, be careful with signs. Remember that subtracting \( (a - \theta) \) from \( 90^\circ \) turns the \( -\theta \) into \( +\theta \).

Question. \( \frac{\tan(45^\circ + \theta) \sec(45^\circ - \theta)}{\cot(45^\circ - \theta) \text{cosec}(45^\circ + \theta)} = 1 \)
Answer: Let us simplify the Left-Hand Side (L.H.S.) of the given identity: \( \text{L.H.S.} = \frac{\tan(45^\circ + \theta) \sec(45^\circ - \theta)}{\cot(45^\circ - \theta) \text{cosec}(45^\circ + \theta)} \) Apply complementary angle transformations to the factors in the numerator: \( \tan(45^\circ + \theta) = \cot[90^\circ - (45^\circ + \theta)] = \cot(45^\circ - \theta) \) \( \sec(45^\circ - \theta) = \text{cosec}[90^\circ - (45^\circ - \theta)] = \text{cosec}(45^\circ + \theta) \) Substitute these into the numerator: \( \text{L.H.S.} = \frac{\cot(45^\circ - \theta) \text{cosec}(45^\circ + \theta)}{\cot(45^\circ - \theta) \text{cosec}(45^\circ + \theta)} = 1 \) This matches the R.H.S., validating the identity.
In simple words: Use complementary formulas to rewrite the top of the fraction. This makes it identical to the bottom, so they divide to give 1.

Exam Tip: Write down your substitution steps clearly. This shows the examiner that you understand the relationship between complementary trigonometric functions.

Question. \( \frac{\sin 54^\circ \times \cos 35^\circ \times \tan 37^\circ}{\cot 53^\circ \times \sin 55^\circ \times \cos 36^\circ} = 1 \)
Answer: Let's simplify the Left-Hand Side (L.H.S.): \( \text{L.H.S.} = \frac{\sin 54^\circ \times \cos 35^\circ \times \tan 37^\circ}{\cot 53^\circ \times \sin 55^\circ \times \cos 36^\circ} \) Let's convert the numerator's trigonometric terms to their complements: \( \sin 54^\circ = \cos(90^\circ - 54^\circ) = \cos 36^\circ \) \( \cos 35^\circ = \sin(90^\circ - 35^\circ) = \sin 55^\circ \) \( \tan 37^\circ = \cot(90^\circ - 37^\circ) = \cot 53^\circ \) Now, substitute these conversions back into the numerator: \( \text{L.H.S.} = \frac{\cos 36^\circ \times \sin 55^\circ \times \cot 53^\circ}{\cot 53^\circ \times \sin 55^\circ \times \cos 36^\circ} \) Rearrange the numerator terms: \( \text{L.H.S.} = \frac{\cot 53^\circ \times \sin 55^\circ \times \cos 36^\circ}{\cot 53^\circ \times \sin 55^\circ \times \cos 36^\circ} = 1 \) L.H.S. equals R.H.S., proving the identity.
In simple words: Change the top terms to their complementary versions. This makes the top and bottom terms the same, which simplifies to 1.

Exam Tip: For expressions with multiple products, look for pairs that add up to \( 90^\circ \) and simplify them systemically to avoid confusion.

Question. \( \tan 6^\circ \tan 12^\circ \tan 84^\circ \tan 78^\circ = 1 \)
Answer: We want to prove that \( \tan 6^\circ \tan 12^\circ \tan 84^\circ \tan 78^\circ = 1 \). First, group the terms with complementary angles together: \( \text{L.H.S.} = (\tan 6^\circ \tan 84^\circ) \times (\tan 12^\circ \tan 78^\circ) \) Using complementary angle relationships, convert one term in each group: \( \tan 84^\circ = \cot(90^\circ - 84^\circ) = \cot 6^\circ \) \( \tan 78^\circ = \cot(90^\circ - 78^\circ) = \cot 12^\circ \) Substitute these back into our expression: \( \text{L.H.S.} = (\tan 6^\circ \cot 6^\circ) \times (\tan 12^\circ \cot 12^\circ) \) Recall that \( \tan \theta \cot \theta = 1 \): \( \text{L.H.S.} = 1 \times 1 = 1 \) L.H.S. equals R.H.S., which verifies the identity.
In simple words: Match the complementary angles. Change one tangent in each pair to cotangent. Since tangent and cotangent are reciprocals, their product is 1.

Exam Tip: Remember the reciprocal identity \( \tan \theta \cdot \cot \theta = 1 \). This is a critical identity that frequently appears in multi-term tangent product questions.

Question. Express \( \sin 6^\circ + \tan 44^\circ \) in terms of t-ratios of angles between \( 45^\circ \) and \( 90^\circ \).
Answer: We have the expression: \( \sin 6^\circ + \tan 44^\circ \) We need to rewrite this expression using angles that fall in the range between \( 45^\circ \) and \( 90^\circ \). Using complementary angle identities: \( \sin \theta = \cos(90^\circ - \theta) \) \( \tan \theta = \cot(90^\circ - \theta) \) Applying these to our terms: \( \sin 6^\circ = \cos(90^\circ - 6^\circ) = \cos 84^\circ \) \( \tan 44^\circ = \cot(90^\circ - 44^\circ) = \cot 46^\circ \) Substituting these values, the expression is rewritten as: \( \cos 84^\circ + \cot 46^\circ \) Both angles, \( 84^\circ \) and \( 46^\circ \), are within the required interval of \( 45^\circ \) to \( 90^\circ \).
In simple words: Swap sine to cosine and tangent to cotangent by subtracting the angles from \( 90^\circ \). This gives you \( 84^\circ \) and \( 46^\circ \), which fit the target range.

Exam Tip: Double-check that your final angles lie strictly within the requested range to avoid losing easy marks.

Question. Express \( \sec 46^\circ + \cot 84^\circ \) in terms of t-ratios of angles between \( 0^\circ \) and \( 45^\circ \).
Answer: Let us consider the expression: \( \sec 46^\circ + \cot 84^\circ \) We need to convert the angles so that they lie in the range between \( 0^\circ \) and \( 45^\circ \). Apply complementary angle relations: \( \sec \theta = \text{cosec}(90^\circ - \theta) \) \( \cot \theta = \tan(90^\circ - \theta) \) Converting each term: \( \sec 46^\circ = \text{cosec}(90^\circ - 46^\circ) = \text{cosec } 44^\circ \) \( \cot 84^\circ = \tan(90^\circ - 84^\circ) = \tan 6^\circ \) Substituting these back into the expression, we get: \( \text{cosec } 44^\circ + \tan 6^\circ \) Since \( 44^\circ \) and \( 6^\circ \) both lie between \( 0^\circ \) and \( 45^\circ \), this is our final answer.
In simple words: Subtract the angles from \( 90^\circ \) while changing secant to cosecant and cotangent to tangent. This gives you smaller angles that fall inside the required range.

Exam Tip: Be careful not to swap the trigonometric ratios incorrectly. Remember that secant pairs with cosecant, and cotangent pairs with tangent.

Question. Find A if \( \cos(A - 33^\circ) = \sin 2A \).
Answer: We are given the equation: \( \cos(A - 33^\circ) = \sin 2A \) Let's convert the Left-Hand Side (L.H.S.) term from cosine to sine using the relation \( \cos \theta = \sin(90^\circ - \theta) \): \( \cos(A - 33^\circ) = \sin[90^\circ - (A - 33^\circ)] \)
\( \implies \cos(A - 33^\circ) = \sin(90^\circ - A + 33^\circ) = \sin(123^\circ - A) \) Substitute this back into the original equation: \( \sin(123^\circ - A) = \sin 2A \) By comparing the angles on both sides: \( 123^\circ - A = 2A \)
\( \implies 3A = 123^\circ \)
\( \implies A = \frac{123^\circ}{3} = 41^\circ \) Thus, the value of A is \( 41^\circ \).
In simple words: Rewrite the cosine term as a sine term by subtracting its angle from \( 90^\circ \). Since both sides are now sines, set the angles equal and solve for A.

Exam Tip: For equations involving two different trigonometric functions, converting one side to match the other is the standard first step to solve for the unknown variable.