Trigonometric Identities MCQ Quiz - Objective Question with Answer for Trigonometric Identities - Download Free PDF

Concept:

\(tan (A+B)= \frac{tanA+tanB}{1-tanA .tanB} \)

Solution:

Given:

If x, y, and z are the angles of a triangle and z = 135°

⇒ x + y + z = 180^o

⇒ x + y = 180^o - 135^o

⇒ x + y = 45o

⇒ tan (x + y) = tan (45o)

⇒ \(\frac{tanx+tany}{1-tanx .tany} = 1\)

⇒ tan x + tan y = 1 - tan x tan y

Adding 1 both side,

⇒ 1 + tan x + tan y = 1 - tan x tan y + 1

⇒ 1 + tan x + tan y + tan x tan y =  2

⇒ 1 + tan x + tan y(1 + tan x) =  2

⇒ (1 + tan x) (1+ tan y) =  2

∴ The value of (1 + tan x) (1 + tan y) is 2

Calculation:

\( \sin z + \cos z = \sin \frac{3\pi}{4} + \cos \frac{3\pi}{4} \)

We can rewrite the terms as:

\( \sin \frac{3\pi}{4} = \sin \left( \pi - \frac{\pi}{4} \right) \) and \( \cos \frac{3\pi}{4} = \cos \left( \pi - \frac{\pi}{4} \right) \)

Using the standard trigonometric identities \(\sin (\pi - \theta) = \sin \theta \) and \(\cos (\pi - \theta) = -\cos \theta \), we get:

\( \sin \frac{3\pi}{4} = \sin \frac{\pi}{4} \) and \( \cos \frac{3\pi}{4} = -\cos \frac{\pi}{4} \)

Now, substitute the values:

\( \sin \frac{\pi}{4} = \frac{1}{\sqrt{2}} \) and \( \cos \frac{\pi}{4} = \frac{1}{\sqrt{2}} \)

Thus, we get:

\( \sin z + \cos z = \frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}} = 0 \)

Hence, the correct answer is Option 1.

Calculation: 

We know:

\( \sin(\alpha) = \frac{3}{5} \quad \text{and} \quad \cos(\alpha) = \frac{4}{5} \)

\( \sin(2\alpha) = 2\sin(\alpha)\cos(\alpha) \)

\( \cos(2\alpha) = 1 - 2\sin^2(\alpha) \)

\( \sin(2\alpha) = 2 \times \frac{3}{5} \times \frac{4}{5} = \frac{24}{25} \)

\( \cos(2\alpha) = 1 - 2 \times \left( \frac{3}{5} \right)^2 = 1 - 2 \times \frac{9}{25} = \frac{7}{25} \)

\( 2\sin(2\alpha) + \cos(2\alpha) = 2 \times \frac{24}{25} + \frac{7}{25} \)

Simplifying:

\( 2\sin(2\alpha) + \cos(2\alpha) = \frac{48}{25} + \frac{7}{25} = \frac{55}{25} = \frac{11}{5} \)

∴ The correct answer is Option (2):

Calculation: 

We are given:

\( 2\sin(\alpha) + \cos(\alpha) = 2 \)

\( \cos(\alpha) = 2(1 - \sin(\alpha)) \)

\( \cos^2(\alpha) = 4(1 - \sin(\alpha))^2 \)

Use the identity \(\cos^2(\alpha) = 1 - \sin^2(\alpha) \)

\( 1 - \sin^2(\alpha) = 4(1 - 2\sin(\alpha) + \sin^2(\alpha)) \)

\( 1 - \sin^2(\alpha) = 4 - 8\sin(\alpha) + 4\sin^2(\alpha) \)

Rearrange the terms to form a quadratic equation:

\( 5\sin^2(\alpha) - 8\sin(\alpha) + 3 = 0 \)

Using the quadratic formula:

\( \sin(\alpha) = \frac{-(-8) \pm \sqrt{(-8)^2 - 4(5)(3)}}{2(5)} \)

\( \sin(\alpha) = \frac{8 \pm \sqrt{64 - 60}}{10} \)

\( \sin(\alpha) = \frac{8 \pm 2}{10} \)

\( \sin(\alpha) = 1 \) or \( \sin(\alpha) = \frac{3}{5} \)

Since \(0^\circ < \alpha < 90^\circ \), we select \(\sin(\alpha) = \frac{3}{5} \)

\( \cos^2(\alpha) = 1 - \sin^2(\alpha) = 1 - \left( \frac{3}{5} \right)^2 = 1 - \frac{9}{25} = \frac{16}{25} \)

\( \cos(\alpha) = \frac{4}{5} \)

\( \tan(\alpha) = \frac{\sin(\alpha)}{\cos(\alpha)} = \frac{\frac{3}{5}}{\frac{4}{5}} = \frac{3}{4} \)

∴ The correct answer is Option (c):

Given:

3sin θ + 5cos θ = 5

Concept:

sin² θ + cos² θ = 1

a² + b² + 2ab = (a + b)²

a2 + b2 - 2ab = (a - b)2

Calculation:

Let 5sin θ - 3cos θ = x    ....(1)

And,

3sin θ + 5cos θ = 5       ....(2)

Squaring in both the equation:

9sin² θ + 25cos² θ + 30sin θ. cos θ= 25                     ....(3)

25sin2 θ + 9cos2 θ - 30sin θ. cos θ  = x²               .....(4)

From equation (3) and equation (4)

⇒ 34 (sin2 θ + cos2 θ) = 25 + x2

⇒ 9 = x2

x = 3 and -3

∴ 5sin θ - 3cos θ = 3

Concept:

cosec² x – cot² x = 1

Calculation:

Given: p = cosec θ – cot θ and q = (cosec θ + cot θ)^-1

⇒ cosec θ + cot θ = 1/q

As we know that, cosec² x – cot² x = 1

⇒ (cosec θ + cot θ) × (cosec θ – cot θ) = 1

\(\frac1q \times p=1\)

Concept:

sin² x + cos² x = 1

Calculation:

Given: sin θ + cos θ = 7/5 

By, squaring both sides of the above equation we get,

⇒ (sin θ + cos θ)² = 49/25

⇒ sin² θ + cos² θ + 2sin θ.cos θ = 49/25

As we know that, sin2 x + cos2 x = 1

⇒ 1 + 2sin θcos θ = 49/25

⇒ 2sin θcos θ = 24/25

Concept:

I. sec² θ – tan² θ = 1

II. a² – b² = (a - b) (a + b)

Calculation:

Given:

tan θ + sec θ = 4     ...(1)

As we know that, sec² θ – tan² θ = 1

⇒ sec² θ – tan² θ = 1

⇒ (sec θ – tan θ) (sec θ + tan θ) = 1

By substituting the value of tan θ + sec θ = 4, in the above equation, we get

⇒ sec θ – tan θ = 1/4     ... (2)

Adding equation (1) and (2), we get

⇒ 2 sec θ = 17/4

⇒ sec θ = 17/8

cos θ = 8/17

Mistake PointsIt is tan θ which is equal to 15/8. The above two equations that we solved were:

tan θ + sec θ = 4

sec θ – tan θ = 1/4

Concept:

a2 - b2 = (a - b) (a + b)

sec2 x - tan2 x = 1

Calculation:

sec4 x - tan4 x

=(sec2 x - tan2 x) (sec2 x + tan2 x)          (∵ a2 - b2 = (a - b) (a + b))

= 1 × (1 + tan2 x + tan2 x)                                          (∵ sec2 x - tan2 x = 1)

= 1 + 2tan² x

Formula used

sin2A + cos2A = 1

1/sin A = cosec A

1/cos A = sec A

Calculation

\(\rm sinA \left(1 + {sinA\over cosA} \right) + cosA \left(1 + {cosA\over sinA} \right) \)

⇒ sin A (cos A + sin A)/cos A + cos A(sin A + cos A)/sin A

⇒ (sin A + cos A) [(sin A/cos A) + (cos A/sin A)]

⇒ (sin A + cos A) [(sin²A + cos²A)/(cos A. sin A)]

⇒ (sin A + cos A) [1/(cos A. sin A)] 

⇒ \( \frac {sin A }{ (cos A. sin A) } + \frac{ cos A }{ (cos A. sin A) } \)

⇒ 1/cos A + 1/sin A

⇒ sec A + Cosec A

The answer is sec A + Cosec A.

Concept:

Trigonometry Formula

sec² θ = 1 + tan² θ

cosec2 θ = 1 + cot2 θ

cot θ = \(\rm \frac{1}{tan \theta }\)

sec θ = \(\rm \frac{1}{cos \theta }\)

cosec θ = \(\rm \frac{1}{sin \theta }\)

Calculation:

\(\rm \frac{1+tan^2\theta}{1+cot^2\theta}-\left(\frac{1-tan\theta}{1-cot\theta}\right)^2\)

\(= \frac{sec^2θ}{cosec^2θ}-\left(\frac{1-tanθ}{1-\frac{1}{tan θ }}\right)^2\)

\(= \rm \frac{\frac{1}{cos^{2}θ}}{\frac{1}{sin^{2}θ}} - \left (\frac{1 - tanθ}{\frac{tan θ - 1}{tan θ }} \right )^{2}\)

\(\rm = \frac{sin^{2}θ }{cos^{2} θ } - \left (\frac{1 - tanθ}{\frac{- (1 - tan θ )}{tan θ }} \right )^{2}\)

= tan² θ - (-tan θ)²

= tan2 θ - tan2 θ 

= 0

Given:

\(\frac{cot^3θ}{cosec^2θ}+\frac{tan^3θ}{sec^2 θ}\) + 2sinθ cosθ 

Formula:

(a + b)² = a² + b² + 2ab

sin²θ + cos²θ = 1

Calculation:

\(\frac{cot^3θ}{cosec^2θ}+\frac{tan^3θ}{sec^2 θ}\) + 2sinθ cosθ 

⇒ \(cos^3θ\over sin^3θ\) × sin²θ + \(sin^3θ\over cos^3θ\) × cos²θ + 2sinθ cosθ 

⇒ \(cos^3θ\over sinθ\) + \(sin^3θ\over cosθ\) + 2sinθ cosθ

⇒ \({sin^4θ+cos^4θ+2sin^2θ cos^2θ}\over{sinθ cosθ}\) 

⇒ \(({sin^2θ+cos^2θ})^2\over{sinθ cosθ}\) = \(1\over{sinθ cosθ}\)

⇒ \({1\over sinθ}\times{1\over cosθ}\) = cosecθ secθ (\(1\over{sinθ}\) = cosecθ, \(1\over{cosθ}\) = secθ) 

∴ \(\frac{cot^3θ}{cosec^2θ}+\frac{tan^3θ}{sec^2 θ}\) + 2sinθ cosθ = cosecθ.secθ

Concept:

• cosec x = 1/sin x
• cot x = cos x/ sin x
• cos² x + sin²x = 1

Calculation:

Here we have to find the value of \(\frac{{\sqrt {1 + {\rm{sinx}}} + \sqrt {1 - {\rm{sinx}}} }}{{\sqrt {1 + {\rm{sinx}}} - \sqrt {1 - {\rm{sinx}}} }}\)

By rationalizing the expression \(\frac{{\sqrt {1 + {\rm{sinx}}} + \sqrt {1 - {\rm{sinx}}} }}{{\sqrt {1 + {\rm{sinx}}} - \sqrt {1 - {\rm{sinx}}} }}\) we get

\(= \frac{{\sqrt {1 + {\rm{sinx}}} + \sqrt {1 - {\rm{sinx}}} }}{{\sqrt {1 + {\rm{sinx}}} - \sqrt {1 - {\rm{sinx}}} }}{\rm{\;}} \times \;\frac{{\sqrt {1 + {\rm{sinx}}} + \sqrt {1 - {\rm{sinx}}} }}{{\sqrt {1 + {\rm{sinx}}} + \sqrt {1 - {\rm{sinx}}} }}\)

\( = \frac{{[\left( {1 + sinx + 1 - sinx} \right) + 2\sqrt {1 - si{n^2}x} }}{{\left( {1 + sinx - 1 + sinx} \right)}}\)

\( = \frac{{2 + 2{\rm{cosx}}}}{{2{\rm{\;sinx}}}}\)

∴ cosec x + cot x

Concept:

sec2 x - tan2 x = 1

Calculation:

Given sec x + tan x = 2     ....(i)

∵ sec² x - tan² x = 1

(sec x + tan x)(sec x - tan x) = 1

2(sec x - tan x) = 1

sec x - tan x = \(1\over 2\)              ....(ii)

Adding the equation (i) and (ii)

2 sec x = 2 + \(1\over 2\)

\(\rm{2\over \cos x} = {5\over 2}\)

cos x = \(4\over 5\)

Given:

√3 - 3√3tan2A = 3tan A - tan3A

Formula used:

tan 3A = (3tan A - tan3A)/(1 - 3tan²A)

Calculation:

√3 - 3√3tan2A = 3tan A - tan3A

⇒ √3(1 - 3tan2A) = 3tan A - tan3A

⇒ √3 = (3tan A - tan3A)/(1 - 3tan2A)

⇒ √3 = tan 3A 

⇒ tan 60° = tan 3A

⇒ 3A = 60° 

⇒ A = 60°/3 = 20° 

∴ The value of A is 20°.