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Introduction to Trigonometry - Trigonometric Identities

Grade 10CBSE

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

🔑Concepts

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Trigonometric identities are equations involving trigonometric ratios of an angle that are true for all values of the angle for which the ratios are defined. The fundamental identity is derived from the Pythagorean theorem applied to a right-angled triangle where a2+b2=c2a^2 + b^2 = c^2. Dividing this equation by different side lengths leads to the three main identities.

Right-angled triangle with hypotenuse 1 showing relationship between sine and cosine.
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The identity sec⁡2θ=1+tan⁡2θ\sec^2 \theta = 1 + \tan^2 \theta is valid for all 0∘≤θ<90∘0^\circ \le \theta < 90^\circ. It links the secant and tangent functions. It can be rearranged as sec⁡2θ−tan⁡2θ=1\sec^2 \theta - \tan^2 \theta = 1 or (sec⁡θ−tan⁡θ)(sec⁡θ+tan⁡θ)=1(\sec \theta - \tan \theta)(\sec \theta + \tan \theta) = 1.

Angular representation of theta for identity domains.
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The identity cosec2θ=1+cot⁡2θ\text{cosec}^2 \theta = 1 + \cot^2 \theta is valid for all 0∘<θ≤90∘0^\circ < \theta \le 90^\circ. This identity is crucial for simplifying expressions involving reciprocal ratios like cosecant and cotangent.

Derivation flow of the third trigonometric identity.
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A common strategy for proving complex identities is to convert all trigonometric ratios (sec, cosec, tan, cot) into sine and cosine terms first, then simplify using algebraic identities like (a+b)2(a+b)^2 or (a2−b2)(a^2 - b^2).

Box highlighting the conversion strategy.

📐Formulae

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

sec⁡2θ−tan⁡2θ=1\sec^2 \theta - \tan^2 \theta = 1

cosec2θ−cot⁡2θ=1\text{cosec}^2 \theta - \cot^2 \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}

sec⁡θ=1cos⁡θ\sec \theta = \frac{1}{\cos \theta}

cosec θ=1sin⁡θ\text{cosec } \theta = \frac{1}{\sin \theta}

💡Examples

Problem 1:

Prove the following identity: cos⁡A1+sin⁡A+1+sin⁡Acos⁡A=2sec⁡A\frac{\cos A}{1 + \sin A} + \frac{1 + \sin A}{\cos A} = 2 \sec A

Solution:

LHS=cos⁡A1+sin⁡A+1+sin⁡Acos⁡A\text{LHS} = \frac{\cos A}{1 + \sin A} + \frac{1 + \sin A}{\cos A} Taking the LCM: =cos⁡2A+(1+sin⁡A)2cos⁡A(1+sin⁡A)= \frac{\cos^2 A + (1 + \sin A)^2}{\cos A (1 + \sin A)} Expanding the numerator: =cos⁡2A+1+sin⁡2A+2sin⁡Acos⁡A(1+sin⁡A)= \frac{\cos^2 A + 1 + \sin^2 A + 2 \sin A}{\cos A (1 + \sin A)} Since sin⁡2A+cos⁡2A=1\sin^2 A + \cos^2 A = 1: =1+1+2sin⁡Acos⁡A(1+sin⁡A)=2+2sin⁡Acos⁡A(1+sin⁡A)= \frac{1 + 1 + 2 \sin A}{\cos A (1 + \sin A)} = \frac{2 + 2 \sin A}{\cos A (1 + \sin A)} Factorizing the numerator: =2(1+sin⁡A)cos⁡A(1+sin⁡A)=2cos⁡A=2sec⁡A=RHS= \frac{2(1 + \sin A)}{\cos A (1 + \sin A)} = \frac{2}{\cos A} = 2 \sec A = \text{RHS}

Explanation:

To solve this, we find a common denominator (LCM), expand the squared term using (a+b)2(a+b)^2, apply the identity sin⁡2A+cos⁡2A=1\sin^2 A + \cos^2 A = 1, and then simplify the fraction by canceling common factors.

Problem 2:

Express the ratio sin⁡A\sin A in terms of sec⁡A\sec A.

Solution:

We know that: sin⁡2A+cos⁡2A=1\sin^2 A + \cos^2 A = 1 Rearranging for sin⁡A\sin A: sin⁡2A=1−cos⁡2A\sin^2 A = 1 - \cos^2 A Since cos⁡A=1sec⁡A\cos A = \frac{1}{\sec A}, we substitute: sin⁡2A=1−1sec⁡2A\sin^2 A = 1 - \frac{1}{\sec^2 A} Taking the LCM on the right side: sin⁡2A=sec⁡2A−1sec⁡2A\sin^2 A = \frac{\sec^2 A - 1}{\sec^2 A} Taking the square root of both sides: sin⁡A=sec⁡2A−1sec⁡A\sin A = \frac{\sqrt{\sec^2 A - 1}}{\sec A}

Explanation:

Using the Pythagorean identity and the reciprocal relationship between cos⁡\cos and sec⁡\sec, we can express one trigonometric ratio purely in terms of another.

Problem 3:

Calculate the value of 9sec⁡2A−9tan⁡2A9 \sec^2 A - 9 \tan^2 A.

Solution:

The expression is: 9sec⁡2A−9tan⁡2A9 \sec^2 A - 9 \tan^2 A Factorizing out 99: 9(sec⁡2A−tan⁡2A)9 (\sec^2 A - \tan^2 A) Using the identity 1+tan⁡2A=sec⁡2A1 + \tan^2 A = \sec^2 A, which implies sec⁡2A−tan⁡2A=1\sec^2 A - \tan^2 A = 1: =9(1)=9= 9(1) = 9

Explanation:

This problem uses a direct application of the secondary trigonometric identity sec⁡2θ−tan⁡2θ=1\sec^2 \theta - \tan^2 \theta = 1 after factoring out the common coefficient.

Problem 4:

Prove the following identity: (cosec θ−cot⁡θ)2=1−cos⁡θ1+cos⁡θ(\text{cosec } \theta - \cot \theta)^2 = \frac{1 - \cos \theta}{1 + \cos \theta}

Conceptual visual of LHS equals RHS balancing.

Solution:

LHS: (cosec θ−cot⁡θ)2(\text{cosec } \theta - \cot \theta)^2 =(1sin⁡θ−cos⁡θsin⁡θ)2= (\frac{1}{\sin \theta} - \frac{\cos \theta}{\sin \theta})^2 =(1−cos⁡θsin⁡θ)2= (\frac{1 - \cos \theta}{\sin \theta})^2 =(1−cos⁡θ)2sin⁡2θ= \frac{(1 - \cos \theta)^2}{\sin^2 \theta} Using sin⁡2θ=1−cos⁡2θ\sin^2 \theta = 1 - \cos^2 \theta: =(1−cos⁡θ)21−cos⁡2θ= \frac{(1 - \cos \theta)^2}{1 - \cos^2 \theta} Using (a2−b2)=(a−b)(a+b)(a^2 - b^2) = (a-b)(a+b): =(1−cos⁡θ)(1−cos⁡θ)(1−cos⁡θ)(1+cos⁡θ)= \frac{(1 - \cos \theta)(1 - \cos \theta)}{(1 - \cos \theta)(1 + \cos \theta)} =1−cos⁡θ1+cos⁡θ=RHS= \frac{1 - \cos \theta}{1 + \cos \theta} = \text{RHS} Hence Proved.

Explanation:

Convert the terms into sine and cosine, common denominator, and use the Pythagorean identity sin⁡2θ+cos⁡2θ=1\sin^2 \theta + \cos^2 \theta = 1 to transform the denominator.

Problem 5:

Prove that: 1+sin⁡A1−sin⁡A=sec⁡A+tan⁡A\sqrt{\frac{1 + \sin A}{1 - \sin A}} = \sec A + \tan A

Symbolic representation of the square root simplification process.

Solution:

LHS: 1+sin⁡A1−sin⁡A\sqrt{\frac{1 + \sin A}{1 - \sin A}} Multiply numerator and denominator by 1+sin⁡A\sqrt{1 + \sin A} (Rationalizing): =(1+sin⁡A)(1+sin⁡A)(1−sin⁡A)(1+sin⁡A)= \sqrt{\frac{(1 + \sin A)(1 + \sin A)}{(1 - \sin A)(1 + \sin A)}} =(1+sin⁡A)21−sin⁡2A= \sqrt{\frac{(1 + \sin A)^2}{1 - \sin^2 A}} Since 1−sin⁡2A=cos⁡2A1 - \sin^2 A = \cos^2 A: =(1+sin⁡A)2cos⁡2A= \sqrt{\frac{(1 + \sin A)^2}{\cos^2 A}} =1+sin⁡Acos⁡A= \frac{1 + \sin A}{\cos A} =1cos⁡A+sin⁡Acos⁡A= \frac{1}{\cos A} + \frac{\sin A}{\cos A} =sec⁡A+tan⁡A=RHS= \sec A + \tan A = \text{RHS} Hence Proved.

Explanation:

Rationalize the denominator inside the square root by multiplying with the conjugate of the denominator, then apply the identity cos⁡2A=1−sin⁡2A\cos^2 A = 1 - \sin^2 A.