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:

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

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.

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:

 $sin\frac{x}{2}$ = $\sqrt{\frac{1-cosx}{2}}$

 $cos\frac{x}{2}$ = $\sqrt{\frac{1+cosx}{2}}$

Calculation:

Given:

cos α and cos β are the roots of the equation  8x2 - 2x - 1 = 0

8x2 - 2x - 1 = 0

⇒ 8x2 - 4x + 2x - 1 = 0

⇒ 4x (2x - 1) + (2x - 1) = 0

⇒ (4x + 1) (2x - 1) = 0

⇒ x =  $-\frac{1}{4}$  and x = $\frac{1}{2}$

Since cos α and cos β are the roots of the equation,

∴ cos α =  $-\frac{1}{4}$  and cos β = $\frac{1}{2}$

⇒  $cos\frac{\alpha }{2}$ = $\sqrt{\frac{1+cos\alpha }{2}}$

⇒ $cos\frac{\alpha }{2}=\sqrt{\frac{1-\frac{1}{4}}{2}}$

⇒ $cos\frac{\alpha }{2}=\sqrt{\frac{3}{8}}=\frac{\sqrt{3}}{2\sqrt{2}}$      ------(1)

and  $cos\frac{\beta }{2}$ = $\sqrt{\frac{1+cos\beta }{2}}$

⇒ $cos\frac{\beta }{2}$ = $\sqrt{\frac{1+\frac{1}{2}}{2}}$

⇒ $cos\frac{\beta }{2}$ = $\frac{\sqrt{3}}{2}$      ------(2)

Multiplying equations (1) and (2), we get

$cos\frac{\alpha }{2}cos\frac{\beta }{2}=\frac{3}{4\sqrt{2}}$

∴​ the value of $cos\frac{\alpha }{2}cos\frac{\beta }{2}=\frac{3}{4\sqrt{2}}$

Concept:

The opposite angles of the cyclic quadrilateral are supplementary.

$\mathrm{\angle }\mathrm{A}+\mathrm{\angle }\mathrm{C}={180}^{\circ }$

$\mathrm{\angle }\mathrm{B}+\mathrm{\angle }\mathrm{D}={180}^{\circ }$

cos(180^∘ - θ) = - cos θ 

cos(180∘ + θ) = - cos θ 

sin(90° - θ) = cosθ 

Calculator:

Statement 1

L.H.S

cos A + cos B

⇒ cos (180^∘ - C) + cos (180^∘ - D)

⇒ - cos C - cos D

⇒ - (cos C + cos D) = R.H.S

Statement 2

L.H.S

cos (π + A) + cos (π - B) + cos (π - C) - sin $(\frac{\pi }{2}-\mathrm{D})$

⇒ - cos A - cos B - cos C - cos D

⇒ - cos (180∘ - C) - cos (180∘ - D) - cos C - cos D

⇒ cos C + cos D - cos C - cos D

⇒ 0 = R.H.S

∴ Both the statements are correct.

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

$\frac{1}{q}\times p=1$

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

$\mathrm{s}\mathrm{i}\mathrm{n}\mathrm{A}(1+\frac{\mathrm{s}\mathrm{i}\mathrm{n}\mathrm{A}}{\mathrm{c}\mathrm{o}\mathrm{s}\mathrm{A}})+\mathrm{c}\mathrm{o}\mathrm{s}\mathrm{A}(1+\frac{\mathrm{c}\mathrm{o}\mathrm{s}\mathrm{A}}{\mathrm{s}\mathrm{i}\mathrm{n}\mathrm{A}})$

⇒ 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)] 

⇒ $\frac{sinA}{(cosA.sinA)}+\frac{cosA}{(cosA.sinA)}$

⇒ 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 = $\frac{1}{\mathrm{t}\mathrm{a}\mathrm{n}\theta }$

sec θ = $\frac{1}{\mathrm{c}\mathrm{o}\mathrm{s}\theta }$

cosec θ = $\frac{1}{\mathrm{s}\mathrm{i}\mathrm{n}\theta }$

Calculation:

$\frac{1+\mathrm{t}\mathrm{a}{\mathrm{n}}^2\theta }{1+\mathrm{c}\mathrm{o}{\mathrm{t}}^2\theta }-{\left(\frac{1-\mathrm{t}\mathrm{a}\mathrm{n}\theta }{1-\mathrm{c}\mathrm{o}\mathrm{t}\theta }\right)}^2$

$=\frac{se{c}^2\theta }{cose{c}^2\theta }-{\left(\frac{1-tan\theta }{1-\frac{1}{tan\theta }}\right)}^2$

$=\frac{\frac{1}{\mathrm{c}\mathrm{o}{\mathrm{s}}^2\theta }}{\frac{1}{\mathrm{s}\mathrm{i}{\mathrm{n}}^2\theta }}-{\left(\frac{1-\mathrm{t}\mathrm{a}\mathrm{n}\theta }{\frac{\mathrm{t}\mathrm{a}\mathrm{n}\theta -1}{\mathrm{t}\mathrm{a}\mathrm{n}\theta }}\right)}^2$

$=\frac{\mathrm{s}\mathrm{i}{\mathrm{n}}^2\theta }{\mathrm{c}\mathrm{o}{\mathrm{s}}^2\theta }-{\left(\frac{1-\mathrm{t}\mathrm{a}\mathrm{n}\theta }{\frac{-(1-\mathrm{t}\mathrm{a}\mathrm{n}\theta )}{\mathrm{t}\mathrm{a}\mathrm{n}\theta }}\right)}^2$

= tan² θ - (-tan θ)²

= tan2 θ - tan2 θ 

= 0

Given:

$\frac{co{t}^3\theta }{cose{c}^2\theta }+\frac{ta{n}^3\theta }{se{c}^2\theta }$ + 2sinθ cosθ 

Formula:

(a + b)² = a² + b² + 2ab

sin²θ + cos²θ = 1

Calculation:

$\frac{co{t}^3\theta }{cose{c}^2\theta }+\frac{ta{n}^3\theta }{se{c}^2\theta }$ + 2sinθ cosθ 

⇒ $\frac{co{s}^3\theta }{si{n}^3\theta }$ × sin²θ + $\frac{si{n}^3\theta }{co{s}^3\theta }$ × cos²θ + 2sinθ cosθ 

⇒ $\frac{co{s}^3\theta }{sin\theta }$ + $\frac{si{n}^3\theta }{cos\theta }$ + 2sinθ cosθ

⇒ $\frac{si{n}^4\theta +co{s}^4\theta +2si{n}^2\theta co{s}^2\theta }{sin\theta cos\theta }$ 

⇒ $\frac{(si{n}^2\theta +co{s}^2\theta {)}^2}{sin\theta cos\theta }$ = $\frac{1}{sin\theta cos\theta }$

⇒ $\frac{1}{sin\theta }\times \frac{1}{cos\theta }$ = cosecθ secθ ($\frac{1}{sin\theta }$ = cosecθ, $\frac{1}{cos\theta }$ = secθ) 

∴ $\frac{co{t}^3\theta }{cose{c}^2\theta }+\frac{ta{n}^3\theta }{se{c}^2\theta }$ + 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 $\frac{\sqrt{1+\mathrm{s}\mathrm{i}\mathrm{n}\mathrm{x}}+\sqrt{1-\mathrm{s}\mathrm{i}\mathrm{n}\mathrm{x}}}{\sqrt{1+\mathrm{s}\mathrm{i}\mathrm{n}\mathrm{x}}-\sqrt{1-\mathrm{s}\mathrm{i}\mathrm{n}\mathrm{x}}}$

By rationalizing the expression $\frac{\sqrt{1+\mathrm{s}\mathrm{i}\mathrm{n}\mathrm{x}}+\sqrt{1-\mathrm{s}\mathrm{i}\mathrm{n}\mathrm{x}}}{\sqrt{1+\mathrm{s}\mathrm{i}\mathrm{n}\mathrm{x}}-\sqrt{1-\mathrm{s}\mathrm{i}\mathrm{n}\mathrm{x}}}$ we get

$=\frac{\sqrt{1+\mathrm{s}\mathrm{i}\mathrm{n}\mathrm{x}}+\sqrt{1-\mathrm{s}\mathrm{i}\mathrm{n}\mathrm{x}}}{\sqrt{1+\mathrm{s}\mathrm{i}\mathrm{n}\mathrm{x}}-\sqrt{1-\mathrm{s}\mathrm{i}\mathrm{n}\mathrm{x}}}\times \frac{\sqrt{1+\mathrm{s}\mathrm{i}\mathrm{n}\mathrm{x}}+\sqrt{1-\mathrm{s}\mathrm{i}\mathrm{n}\mathrm{x}}}{\sqrt{1+\mathrm{s}\mathrm{i}\mathrm{n}\mathrm{x}}+\sqrt{1-\mathrm{s}\mathrm{i}\mathrm{n}\mathrm{x}}}$

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

$=\frac{2+2\mathrm{c}\mathrm{o}\mathrm{s}\mathrm{x}}{2\mathrm{s}\mathrm{i}\mathrm{n}\mathrm{x}}$

∴ 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 = $\frac{1}{2}$              ....(ii)

Adding the equation (i) and (ii)

2 sec x = 2 + $\frac{1}{2}$

$\frac{2}{\cos \mathrm{x}}=\frac{5}{2}$

cos x = $\frac{4}{5}$

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°.