krit.club logo

Trigonometric Identities and Applications - Establish and use simple trigonometric identities based on fundamental ratio relations

Grade 10CBSE

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

🔑Concepts

•

The fundamental trigonometric identities are derived from the unit circle and the Pythagorean theorem applied to a right-angled triangle where the hypotenuse is 1 unit. For any angle θ\theta in a right triangle, the coordinates of a point on the unit circle are (cos⁡θ,sin⁡θ)(\cos \theta, \sin \theta).

Unit circle demonstrating the relationship between coordinates and sine/cosine.
•

Reciprocal relations link the primary ratios: sin⁡θ\sin \theta is the reciprocal of csc⁡θ\csc \theta, cos⁡θ\cos \theta is the reciprocal of sec⁡θ\sec \theta, and tan⁡θ\tan \theta is the reciprocal of cot⁡θ\cot \theta.

•

Quotient relations express tan⁡θ\tan \theta and cot⁡θ\cot \theta 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}.

•

The identity sin⁡2θ+cos⁡2θ=1\sin^2 \theta + \cos^2 \theta = 1 is the most fundamental identity and is used to transform expressions involving squares of sine and cosine.

•

The identities 1+tan⁡2θ=sec⁡2θ1 + \tan^2 \theta = \sec^2 \theta and 1+cot⁡2θ=csc⁡2θ1 + \cot^2 \theta = \csc^2 \theta relate the squared values of reciprocal functions.

📐Formulae

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

1+tan⁡2θ=sec⁡2θ1 + \tan^2 \theta = \sec^2 \theta

1+cot⁡2θ=csc⁡2θ1 + \cot^2 \theta = \csc^2 \theta

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}

csc⁡θ=1sin⁡θ\csc \theta = \frac{1}{\sin \theta}

💡Examples

Problem 1:

Prove the identity: (sec⁡A+tan⁡A)(1−sin⁡A)=cos⁡A(\sec A + \tan A)(1 - \sin A) = \cos A.

Solution:

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

Explanation:

The problem is solved by expressing sec⁡A\sec A and tan⁡A\tan A in terms of cos⁡A\cos A and sin⁡A\sin A, then applying the algebraic identity (a+b)(a−b)=a2−b2(a+b)(a-b) = a^2 - b^2 followed by the Pythagorean identity sin⁡2A+cos⁡2A=1\sin^2 A + \cos^2 A = 1.

Problem 2:

Prove that sin⁡θ1+cos⁡θ+1+cos⁡θsin⁡θ=2csc⁡θ\frac{\sin \theta}{1 + \cos \theta} + \frac{1 + \cos \theta}{\sin \theta} = 2\csc \theta.

Solution:

LHS =sin⁡2θ+(1+cos⁡θ)2sin⁡θ(1+cos⁡θ)= \frac{\sin^2 \theta + (1 + \cos \theta)^2}{\sin \theta(1 + \cos \theta)} =sin⁡2θ+1+cos⁡2θ+2cos⁡θsin⁡θ(1+cos⁡θ)= \frac{\sin^2 \theta + 1 + \cos^2 \theta + 2\cos \theta}{\sin \theta(1 + \cos \theta)} Since sin⁡2θ+cos⁡2θ=1\sin^2 \theta + \cos^2 \theta = 1: =1+1+2cos⁡θsin⁡θ(1+cos⁡θ)= \frac{1 + 1 + 2\cos \theta}{\sin \theta(1 + \cos \theta)} =2+2cos⁡θsin⁡θ(1+cos⁡θ)= \frac{2 + 2\cos \theta}{\sin \theta(1 + \cos \theta)} =2(1+cos⁡θ)sin⁡θ(1+cos⁡θ)= \frac{2(1 + \cos \theta)}{\sin \theta(1 + \cos \theta)} =2sin⁡θ=2csc⁡θ== \frac{2}{\sin \theta} = 2\csc \theta = RHS.

Explanation:

We first take the LCM of the denominators. Then we expand the square and use the fundamental identity sin⁡2θ+cos⁡2θ=1\sin^2 \theta + \cos^2 \theta = 1. Finally, we factorize the numerator to cancel the common term (1+cos⁡θ)(1 + \cos \theta).

Problem 3:

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

Right triangle illustrating the relationship between sine and cosine.

Solution:

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

Explanation:

To solve expressions involving square roots, rationalize the denominator or numerator to create perfect squares, then apply the Pythagorean identity 1−sin⁡2A=cos⁡2A1 - \sin^2 A = \cos^2 A.

Problem 4:

Show that cos⁡A1−tan⁡A+sin⁡A1−cot⁡A=sin⁡A+cos⁡A\frac{\cos A}{1 - \tan A} + \frac{\sin A}{1 - \cot A} = \sin A + \cos A.

Graph showing the behavior of tan(A) used in the identity.

Solution:

LHS = cos⁡A1−sin⁡Acos⁡A+sin⁡A1−cos⁡Asin⁡A\frac{\cos A}{1 - \frac{\sin A}{\cos A}} + \frac{\sin A}{1 - \frac{\cos A}{\sin A}} LHS = cos⁡Acos⁡A−sin⁡Acos⁡A+sin⁡Asin⁡A−cos⁡Asin⁡A\frac{\cos A}{\frac{\cos A - \sin A}{\cos A}} + \frac{\sin A}{\frac{\sin A - \cos A}{\sin A}} LHS = cos⁡2Acos⁡A−sin⁡A+sin⁡2Asin⁡A−cos⁡A\frac{\cos^2 A}{\cos A - \sin A} + \frac{\sin^2 A}{\sin A - \cos A} Take a negative sign common from the second denominator: LHS = cos⁡2Acos⁡A−sin⁡A−sin⁡2Acos⁡A−sin⁡A\frac{\cos^2 A}{\cos A - \sin A} - \frac{\sin^2 A}{\cos A - \sin A} LHS = cos⁡2A−sin⁡2Acos⁡A−sin⁡A\frac{\cos^2 A - \sin^2 A}{\cos A - \sin A} Use the identity a2−b2=(a−b)(a+b)a^2 - b^2 = (a - b)(a + b): LHS = (cos⁡A−sin⁡A)(cos⁡A+sin⁡A)cos⁡A−sin⁡A\frac{(\cos A - \sin A)(\cos A + \sin A)}{\cos A - \sin A} LHS = cos⁡A+sin⁡A\cos A + \sin A LHS = RHS

Explanation:

Convert all trigonometric ratios to sine and cosine. Simplify the complex fractions and use algebraic factorization to cancel common terms.