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Trigonometry - Trigonometrical Identities

Grade 10ICSE

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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The fundamental trigonometric ratios are defined based on a right-angled triangle. For an angle θ\theta, the side opposite to it is the perpendicular (PP), the side adjacent is the base (BB), and the longest side is the hypotenuse (HH).

Right-angled triangle showing theta, perpendicular, base, and hypotenuse.
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The reciprocal identities connect sine with cosecant, cosine with secant, and tangent with cotangent. Multiplying a ratio by its reciprocal always equals 1, such as sin⁡θ⋅csc⁡θ=1\sin \theta \cdot \csc \theta = 1.

Flowchart showing reciprocal relationships between trigonometric functions.
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The Pythagorean Identity sin⁡2θ+cos⁡2θ=1\sin^2 \theta + \cos^2 \theta = 1 is derived from the Pythagorean theorem a2+b2=c2a^2 + b^2 = c^2. On a unit circle (radius 1), the coordinates of any point (x,y)(x, y) are (cos⁡θ,sin⁡θ)(\cos \theta, \sin \theta).

Unit circle showing point P with coordinates cos theta and sin theta.
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Quotient identities express tangent and cotangent in terms of sine and cosine: tan⁡θ=sin⁡θcos⁡θ\tan \theta = \frac{\sin \theta}{\cos \theta} and cot⁡θ=cos⁡θsin⁡θ\cot \theta = \frac{\cos \theta}{\sin \theta}. These are essential for simplifying complex expressions.

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To prove identities, start from the more complex side (usually LHS) and use algebraic techniques like taking LCM, rationalization, or factoring alongside basic identities to reach the simpler side.

📐Formulae

sin⁡θ=1csc⁡θ  ⟹  sin⁡θ⋅csc⁡θ=1\sin \theta = \frac{1}{\csc \theta} \implies \sin \theta \cdot \csc \theta = 1

cos⁡θ=1sec⁡θ  ⟹  cos⁡θ⋅sec⁡θ=1\cos \theta = \frac{1}{\sec \theta} \implies \cos \theta \cdot \sec \theta = 1

tan⁡θ=1cot⁡θ  ⟹  tan⁡θ⋅cot⁡θ=1\tan \theta = \frac{1}{\cot \theta} \implies \tan \theta \cdot \cot \theta = 1

tan⁡θ=sin⁡θcos⁡θ\tan \theta = \frac{\sin \theta}{\cos \theta}

cot⁡θ=cos⁡θsin⁡θ\cot \theta = \frac{\cos \theta}{\sin \theta}

sin⁡2θ+cos⁡2θ=1\sin^2 \theta + \cos^2 \theta = 1

sin⁡2θ=1−cos⁡2θ\sin^2 \theta = 1 - \cos^2 \theta

cos⁡2θ=1−sin⁡2θ\cos^2 \theta = 1 - \sin^2 \theta

1+tan⁡2θ=sec⁡2θ  ⟹  sec⁡2θ−tan⁡2θ=11 + \tan^2 \theta = \sec^2 \theta \implies \sec^2 \theta - \tan^2 \theta = 1

1+cot⁡2θ=csc⁡2θ  ⟹  csc⁡2θ−cot⁡2θ=11 + \cot^2 \theta = \csc^2 \theta \implies \csc^2 \theta - \cot^2 \theta = 1

💡Examples

Problem 1:

Prove that sin⁡A1+cos⁡A+1+cos⁡Asin⁡A=2csc⁡A\frac{\sin A}{1 + \cos A} + \frac{1 + \cos A}{\sin A} = 2\csc A

Solution:

Step 1: Take the LHS and find the common denominator. LHS=sin⁡2A+(1+cos⁡A)2sin⁡A(1+cos⁡A)LHS = \frac{\sin^2 A + (1 + \cos A)^2}{\sin A(1 + \cos A)} Step 2: Expand the term (1+cos⁡A)2(1 + \cos A)^2 using the identity (a+b)2=a2+2ab+b2(a+b)^2 = a^2 + 2ab + b^2. LHS=sin⁡2A+1+2cos⁡A+cos⁡2Asin⁡A(1+cos⁡A)LHS = \frac{\sin^2 A + 1 + 2\cos A + \cos^2 A}{\sin A(1 + \cos A)} Step 3: Group sin⁡2A\sin^2 A and cos⁡2A\cos^2 A together. LHS=(sin⁡2A+cos⁡2A)+1+2cos⁡Asin⁡A(1+cos⁡A)LHS = \frac{(\sin^2 A + \cos^2 A) + 1 + 2\cos A}{\sin A(1 + \cos A)} Step 4: Use the identity sin⁡2A+cos⁡2A=1\sin^2 A + \cos^2 A = 1. LHS=1+1+2cos⁡Asin⁡A(1+cos⁡A)=2+2cos⁡Asin⁡A(1+cos⁡A)LHS = \frac{1 + 1 + 2\cos A}{\sin A(1 + \cos A)} = \frac{2 + 2\cos A}{\sin A(1 + \cos A)} Step 5: Factor out 2 from the numerator. LHS=2(1+cos⁡A)sin⁡A(1+cos⁡A)LHS = \frac{2(1 + \cos A)}{\sin A(1 + \cos A)} Step 6: Cancel the common factor (1+cos⁡A)(1 + \cos A). LHS=2sin⁡A=2csc⁡A=RHSLHS = \frac{2}{\sin A} = 2\csc A = RHS

Explanation:

This problem is solved by using the algebraic technique of finding a common denominator and then applying the fundamental Pythagorean identity sin⁡2θ+cos⁡2θ=1\sin^2 \theta + \cos^2 \theta = 1 to simplify the numerator.

Problem 2:

Prove that 1−cos⁡θ1+cos⁡θ=csc⁡θ−cot⁡θ\sqrt{\frac{1 - \cos \theta}{1 + \cos \theta}} = \csc \theta - \cot \theta

Solution:

Step 1: Start with the LHS and rationalize the denominator by multiplying the numerator and denominator by (1−cos⁡θ)(1 - \cos \theta) inside the square root. LHS=(1−cos⁡θ)(1−cos⁡θ)(1+cos⁡θ)(1−cos⁡θ)LHS = \sqrt{\frac{(1 - \cos \theta)(1 - \cos \theta)}{(1 + \cos \theta)(1 - \cos \theta)}} Step 2: Simplify the numerator and denominator. LHS=(1−cos⁡θ)21−cos⁡2θLHS = \sqrt{\frac{(1 - \cos \theta)^2}{1 - \cos^2 \theta}} Step 3: Use the identity 1−cos⁡2θ=sin⁡2θ1 - \cos^2 \theta = \sin^2 \theta. LHS=(1−cos⁡θ)2sin⁡2θLHS = \sqrt{\frac{(1 - \cos \theta)^2}{\sin^2 \theta}} Step 4: Remove the square root. LHS=1−cos⁡θsin⁡θLHS = \frac{1 - \cos \theta}{\sin \theta} Step 5: Split the fraction. LHS=1sin⁡θ−cos⁡θsin⁡θLHS = \frac{1}{\sin \theta} - \frac{\cos \theta}{\sin \theta} Step 6: Apply reciprocal and quotient identities. LHS=csc⁡θ−cot⁡θ=RHSLHS = \csc \theta - \cot \theta = RHS

Explanation:

The key strategy here is 'rationalizing' the expression under the square root to create perfect squares in both the numerator and denominator, allowing the root to be removed.

Problem 3:

Prove that (sec⁡A−tan⁡A)2=1−sin⁡A1+sin⁡A(\sec A - \tan A)^2 = \frac{1 - \sin A}{1 + \sin A}.

Logical flow for proving the identity by converting to sine and cosine.

Solution:

LHS: (sec⁡A−tan⁡A)2(\sec A - \tan A)^2 Converting to sin⁡\sin and cos⁡\cos: =(1cos⁡A−sin⁡Acos⁡A)2= (\frac{1}{\cos A} - \frac{\sin A}{\cos A})^2 =(1−sin⁡Acos⁡A)2= (\frac{1 - \sin A}{\cos A})^2 =(1−sin⁡A)2cos⁡2A= \frac{(1 - \sin A)^2}{\cos^2 A} Using cos⁡2A=1−sin⁡2A\cos^2 A = 1 - \sin^2 A: =(1−sin⁡A)21−sin⁡2A= \frac{(1 - \sin A)^2}{1 - \sin^2 A} =(1−sin⁡A)(1−sin⁡A)(1−sin⁡A)(1+sin⁡A)= \frac{(1 - \sin A)(1 - \sin A)}{(1 - \sin A)(1 + \sin A)} =1−sin⁡A1+sin⁡A=RHS= \frac{1 - \sin A}{1 + \sin A} = RHS

Explanation:

This example demonstrates the strategy of converting secant and tangent into sine and cosine terms first. Then, the identity cos⁡2A=1−sin⁡2A\cos^2 A = 1 - \sin^2 A is used to factorize the denominator to cancel out common terms.

Problem 4:

Prove that 11+sin⁡θ+11−sin⁡θ=2sec⁡2θ\frac{1}{1 + \sin \theta} + \frac{1}{1 - \sin \theta} = 2\sec^2 \theta.

Box highlighting the key identity substitution needed for the proof.

Solution:

LHS: 11+sin⁡θ+11−sin⁡θ\frac{1}{1 + \sin \theta} + \frac{1}{1 - \sin \theta} Taking LCM: =(1−sin⁡θ)+(1+sin⁡θ)(1+sin⁡θ)(1−sin⁡θ)= \frac{(1 - \sin \theta) + (1 + \sin \theta)}{(1 + \sin \theta)(1 - \sin \theta)} =21−sin⁡2θ= \frac{2}{1 - \sin^2 \theta} Using sin⁡2θ+cos⁡2θ=1  ⟹  1−sin⁡2θ=cos⁡2θ\sin^2 \theta + \cos^2 \theta = 1 \implies 1 - \sin^2 \theta = \cos^2 \theta: =2cos⁡2θ= \frac{2}{\cos^2 \theta} =2×1cos⁡2θ= 2 \times \frac{1}{\cos^2 \theta} =2sec⁡2θ=RHS= 2\sec^2 \theta = RHS

Explanation:

To solve this, we combine the fractions by finding a common denominator. The denominator results in a difference of squares (1−sin⁡2θ)(1 - \sin^2 \theta), which is transformed into cos⁡2θ\cos^2 \theta using the primary Pythagorean identity.