Trigonometric Identities Practice Problems

Last Updated : 18 Jun, 2026

Trigonometric identities are a set of formulas that can be used to reduce a variety of complex equations that contain trigonometric functions.

  • These identities connect the various trigonometric functions—sine (sin), cosine (cos), tangent (tan), and their reciprocals (cotangent, secant, and cosecant).

Problem 1: Find the value of \frac{\sin x}{1 + \cos x} + \frac{\cos x}{1 + \sin x}.

To simplify this expression, we can find a common denominator:

\frac{\sin x(1+\sin x)+ \cos x(1+\cos x)}{(1+\cos x)(1+\sin x)}

Expanding the numerator, we get

sin x + sin2 x + cos x + cos2 x

As we know sin2 x + cos2 x = 1, hence the above equation becomes:

sin x + cos x + 1

Hence the value of given expression is: \frac{sin x + cos x + 1} {(1 + cos x)(1+sin x)}

Problem 2: Prove that sin (45° – a) cos (45° – b) + cos (45° – a) sin (45° – b)  = cos (a + b).

Let us solve the LHS of the given equation: 

By using formula: sin (A + B) = sin A cos B + cos A sin B we get

sin(45° – a) cos (45° – b) + cos (45° – a) sin (45° – b) = sin [(45°– a) + (45° – b)]

= sin [90° – (a + b)]

As sin (90° – θ) = cos θ, hence

sin [90° – (a + b)] = cos (a + b)

= R. H. S  

∴ LHS = RHS [Hence Proved]

Problem 3: Show that (tan2 θ + tan4 θ) = (sec4 θ – sec2 θ)

Let us take the RHS of the given equation:

We have sec4θ – sec2θ

Take sec2θ common

sec2θ(sec2θ – 1)

We know, sec2θ = 1 + tan2θ, Hence the above equation become:

(1 + tan2θ) (1 + tan2θ – 1)

⇒ (1 + tan2θ) tan2θ

⇒ (tan2θ + tan4θ) = LHS      

∴ LHS = RHS [Hence Proved]

Problem 4: Find the value of sin(π/4 - π/6)

Given, sin (π/4 - π/6)

By using formula: sin (A – B) = sin A cos B – cos A sin B, we get

sin (π/4 - π/6) = sin π/4 cos π/6 – cos π/4 sin π/6   

Since, cos π/4 = sin π/4 = 1/√2, cos π/6 = √3/2, and sin π/6 = 1/2

Putting these values above we get,

sin (π/4 - π/6) = (1/√2) (√3/2) – (1/√2)(1/2)

= (√3 – 1)/2√2

Hence, sin (π/4 - π/6) = (√3 – 1)/2√2

Problem 5: Solve (1 + tan2θ) cos2θ  

Given, (1 + tan2θ)cos2θ

Since we know 1 + tan2θ = sec2θ Hence the above equation becomes:

sec2θ . cos2θ

(1/cos2θ) . cos2θ = 1

Hence (1 + tan2θ)cos2θ = 1

Practice Problems

P1. Simplify the expression \frac{sin^2}{1-cosx} + \frac{cos^2}{1+sinx}.

P2. Prove the identity \frac1{sinx \cdot cosx} = \frac1{sinx} + \frac 1{cosx}

P3. Prove the identity \frac {\tan x} {1-\cot x} + \frac {\cot x} {1-\tan x}

P4. Simplify the expression \frac {\sin^2 x}{\cos x} +\frac {\cos^2 x}{\sin x}

P5. Prove the identity sinx tanx + cosx cotx = 2.

P6. Simplify the expression \frac 1{\cos x} \cdot \frac1{1+ \sin x} +\frac 1{\sin x} \cdot \frac1{1+ \cos x}

P7. Evaluate: \frac {1 + \tan x}{1 - \tan x} = \frac {1 + \sin x}{1 - \sin x}

P8. Prove the identity sin2 x + cos2 x = 1

P9. Prove the identity \frac{\sin x} {1-\cos x} + \frac{\cos x} {1-\sin x} = \frac 2 {\sin x+ \cos x}

P10. Simplify the expression \sin x + \tan x \cdot \cos x + \cot x \cdot \cos x

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