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

Calculation: 

We are given:

\( p = \tan(2\alpha) - \tan(\alpha) \)

\( q = \cot(\alpha) - \cot(2\alpha) \)

We need to find \(\tan^2(\alpha) \) in terms of p  and q .

We start by simplifying the expression \(\frac{p}{p + 2q} \)

\( \frac{p}{p + 2q} = \frac{1}{1 + \frac{2q}{p}} \)

This simplifies further as:

\( \frac{1}{1 + \frac{2}{\tan(\alpha) \cdot \tan(2\alpha)}} \)

Now, simplify the fraction inside the denominator:

\( = \frac{1}{1 + \frac{2}{\tan(\alpha) \cdot \tan(2\alpha)}} = \frac{1}{1 + \frac{1}{2} \cdot \tan(\alpha) \cdot \tan(2\alpha)} \)

Now, expand both terms using trigonometric identities:

\( = \frac{\sin(\alpha) \cdot \sin(2\alpha)}{\sin(\alpha) \cdot \cos(\alpha) + \cos(\alpha) \cdot \cos(2\alpha)} \)

Simplifying this expression gives:

\( = \tan^2(\alpha) \)

∴ The correct answer is Option (c):

Calculation: 

We are given:

\( p = \tan(2\alpha) - \tan(\alpha) \)

\( q = \cot(\alpha) - \cot(2\alpha) \)

We need to find p + q.

\( p + q = \left( \frac{\sin^2(2\alpha) - \cos^2(2\alpha)}{\sin(2\alpha) \cdot \cos(2\alpha)} \right) + \left( \frac{\cos^2(\alpha) - \sin^2(\alpha)}{\sin(\alpha) \cdot \cos(\alpha)} \right) \)

Simplifying both terms:

\( = \frac{-2\cos(4\alpha)}{\sin(4\alpha)} + \frac{2\cos(2\alpha)}{\sin(2\alpha)} \)

Now, factorizing and simplifying further:

\( = \frac{2(\sin(4\alpha) \cdot \cos(2\alpha) - \cos(4\alpha) \cdot \sin(2\alpha))}{\sin(4\alpha) \cdot \sin(2\alpha)} \)

Recognizing the sine identity, we get:

\( = \frac{2 \sin(4\alpha - 2\alpha)}{\sin(4\alpha) \cdot \sin(2\alpha)} \)

Finally, simplifying this:

\( = \frac{2 \sin(2\alpha)}{\sin(4\alpha) \cdot \sin(2\alpha)} \)

The final result is:

\( p + q = 2 cosec (4\alpha) \)

∴ The correct answer is Option (4)

Explanation:

We are given:

\( p = \tan(2\alpha) - \tan(\alpha) \)

\( q = \cot(\alpha) - \cot(2\alpha) \)

We rewrite q  in terms of tangent since \(\cot(\theta) = \frac{1}{\tan(\theta)} \)

\( q = \frac{1}{\tan(\alpha)} - \frac{1}{\tan(2\alpha)} \)

Now we compute \(\cot(\theta) = \frac{1}{\tan(\theta)} \)

\( \frac{p}{q} = \frac{\tan(2\alpha) - \tan(\alpha)}{\frac{1}{\tan(\alpha)} - \frac{1}{\tan(2\alpha)}} \)

Next, we find a common denominator for q :

\( \frac{1}{\tan(\alpha)} - \frac{1}{\tan(2\alpha)} = \frac{\tan(2\alpha) - \tan(\alpha)}{\tan(\alpha) \cdot \tan(2\alpha)} \)

Substituting this back into the formula for \(\frac{p}{q} \)

\( \frac{p}{q} = \frac{\tan(2\alpha) - \tan(\alpha)}{\frac{\tan(2\alpha) - \tan(\alpha)}{\tan(\alpha) \cdot \tan(2\alpha)}} \)

Simplifying, we get:

\( \frac{p}{q} = \tan(\alpha) \cdot \tan(2\alpha) \)

∴ The correct answer is Option (c): \( \tan(\alpha) \cdot \tan(2\alpha) \)

We are given the following equations:

\[ a \cos \theta + b \sin \theta = c \]

\[ a \sin \theta - b \cos \theta = d \]

To find a relationship between \( a \), \( b \), \( c \), and \( d \), we square both equations and add them:

\[ (a \cos \theta + b \sin \theta)^2 + (a \sin \theta - b \cos \theta)^2 = c^2 + d^2 \]

Expanding the left-hand side:

\[ a^2 \cos^2 \theta + 2ab \cos \theta \sin \theta + b^2 \sin^2 \theta + a^2 \sin^2 \theta - 2ab \cos \theta \sin \theta + b^2 \cos^2 \theta = c^2 + d^2 \]

Simplifying:

\[ a^2 (\cos^2 \theta + \sin^2 \theta) + b^2 (\sin^2 \theta + \cos^2 \theta) = c^2 + d^2 \]

Using the Pythagorean identity \( \cos^2 \theta + \sin^2 \theta = 1 \):

\[ a^2 + b^2 = c^2 + d^2 \]

Thus, the relationship between \( a \), \( b \), \( c \), and \( d \) is:

\[ \boxed{a^2 + b^2 = c^2 + d^2} \]

To evaluate the expression:

\[ \sin^2 30^\circ + \cos^2 45^\circ + \sin^2 60^\circ + \cos^2 120^\circ + \sin^2 150^\circ \]

we calculate each term individually:

\[ \sin^2 30^\circ = \left(\frac{1}{2}\right)^2 = \frac{1}{4} \]

\[ \cos^2 45^\circ = \left(\frac{1}{\sqrt{2}}\right)^2 = \frac{1}{2} \]

\[ \sin^2 60^\circ = \left(\frac{\sqrt{3}}{2}\right)^2 = \frac{3}{4} \]

\[ \cos^2 120^\circ = \left(-\frac{1}{2}\right)^2 = \frac{1}{4} \]

\[ \sin^2 150^\circ = \left(\frac{1}{2}\right)^2 = \frac{1}{4} \]

Adding all the terms:

\[ \frac{1}{4} + \frac{1}{2} + \frac{3}{4} + \frac{1}{4} + \frac{1}{4} = 2 \]

Thus, the value of the expression is:

\[ \boxed{2} \]

Concept:

sin2x + cos2x = 1

Calculation:

Given: sin x + sin² x = 1

As we know that, sin2x + cos2x = 1

⇒ sin x + (1 – cos² x) = 1

⇒ sin x = cos² x ----(1)

⇒ sin² x = cos⁴ x (Using (1)) 

Given:

(1 + cos²A) = 3sinA.cosA

Solution:

Dividing by cos²A,

⇒ (1 + cos2A) = 3sinA.cosA

⇒ (1 + sec²A) = 3tanA

We know that: sec²A = (1 + tan²A)

⇒ 2 + tan²A = 3tanA

⇒ tan²A – 3tanA + 2 = 0

⇒ (tan A – 1) (tan A – 2) = 0

⇒ tan A = 1 or 2

Calculation:

Given tan θ = \(-\frac{4}{3}\).

Consider a right angled triangle with perpendicular 4 units and base 3 units.

By pythagorus theorem h = \(\rm\sqrt{p^2+b^2} \)

⇒ h = 5.

∴ sin θ = \(\rm\frac{p}{h}\)

⇒ sin θ = \(\frac{4}{5}\)

We know that tan function is negative in 2nd and 4th quadrant.

Sin function is positive in 2nd quadrant and negative in 4th quadrant.

If θ is in 2nd quadrant, sin θ  = \(\rm\frac{4}{5}\).

If θ is in 4th quadrant, sin θ = \(\frac{-4}{5}\).

sin θ can be \(\frac{4}{5}\) or \(-\frac{4}{5}\).

Concept:

• sin² x + cos² x = 1
• cos (A + B) = cos A cos B – sin A sin B
• cos (A – B) = cos A cos B + sin A sin B

Calculation:

Given: \(\cos \theta = \frac{8}{{17}}\) and θ is in 1^st quadrant

\(\sin \theta = \sqrt {1 - {{\cos }^2}\theta } = \frac{{15}}{{17}}\)

cos (30 + θ) = cos 30 cos θ – sin 30 sin θ

\(= \frac{{\sqrt 3 }}{2}\left( {\frac{8}{{17}}} \right) - \left( {\frac{1}{2}} \right)\left( {\frac{{15}}{{17}}} \right)\)

\(= {{8\sqrt 3 - 15}}{{34}}\)     ----(1)

cos (45 – θ) = cos 45 cos θ + sin 45 sin θ

\(= \left( {\frac{1}{{\sqrt 2 }}} \right)\left( {\frac{8}{{17}}} \right) + \left( {\frac{1}{{\sqrt 2 }}} \right)\left( {\frac{{15}}{{17}}} \right)\)

\(= \frac{{23\sqrt 2}}{{34}}\)      ----(2)

cos (120 – θ) = cos 120 cos θ + sin 120 sin θ

\(= \left( { - \frac{1}{2}} \right)\left( {\frac{8}{{17}}} \right) + \left( {\frac{{\sqrt 3 }}{2}} \right)\left( {\frac{{15}}{{17}}} \right)\)      

\(= \frac{{15\sqrt 3 - 8}}{{34}}\) ----(3)

Adding (1), (2) and (3)

\(\frac{{8\sqrt 3 - 15}}{{34}} + \frac{{23\sqrt 2 }}{{34}} + \frac{{15\sqrt 3 - 8}}{{34}}\)

\(\Rightarrow \frac{{23\sqrt 3 + 23\sqrt 2 - 23}}{{34}}\)

Given

(cosec x + cot x)/(cosec x – cot x) = 7

Formula used

Cos²x = 1 – sin²x

Calculation

(cosec x + cot x)/(cosec x – cot x) = 7

⇒ (cosec x + cot x = 7(cosec x – cot x)

⇒ 8cot x = 6 cosec x

⇒ (8cos x)/sin x = 6/sin x

⇒ Cos x = 3/4

⇒ Cos² x = 9/16

⇒ Sin² x = 1 – 9/16

⇒ Sin² x = 7/16

⇒ (4sin2x + 1)/(4sin2x – 1) = (4 × 7/16 + 1)/(4 × 7/16 – 1)

⇒ (7/4 + 1)/(7/4 –1)

⇒ (11/4)/(3/4)

⇒ 11/3

∴ The value of  (4sin2x + 1)/(4sin2x – 1) is 11/3

\(\sin \theta = \frac{{{p^2} - 1}}{{{p^2} + 1}}\)

∴ sin² θ + cos² θ = 1

\(\Rightarrow {\cos ^2}\theta = 1 - \frac{{{{\left( {{p^2} - 1} \right)}^2}}}{{{{\left( {{p^2} + 1} \right)}^2}}} = \frac{{{{\left( {{p^2} + 1} \right)}^2} - {{\left( {{p^2} - 1} \right)}^2}}}{{{{\left( {{p^2} + 1} \right)}^2}}}\)

\(\Rightarrow {\cos ^2}\theta = \frac{{4{p^2}}}{{{{\left( {{p^2} + 1} \right)}^2}}}\)

Concept:

sec^-1 x = cos^-1 (1/x)

cosec^-1 (x) + sec^-1 (x) = π / 2

Calculation:

As we know that, 

sec-1 x = cos-1 (1/x)

\({\rm{co}}{{\rm{s}}^{ - 1}}\left( {\frac{1}{{\sqrt 5 }}} \right) = {\rm{\theta }} \)

\(\Rightarrow {\rm{se}}{{\rm{c}}^{ - 1}}\left( {\sqrt 5 } \right) = {\rm{\theta }}\)

As we know that,

cosec-1 (x) + sec-1 (x) = π / 2

\(\Rightarrow \frac{{\rm{\pi }}}{2} - {\rm{cose}}{{\rm{c}}^{ - 1}}\left( {\sqrt 5 } \right) = {\rm{\theta }}\)

\(\therefore {\rm{cose}}{{\rm{c}}^{ - 1}}\left( {\sqrt 5 } \right) = \frac{{\rm{\pi }}}{2} - {\rm{\theta }}\)

Given:

tan \(A=\frac{{1.1}}{{6}}\)

⇒ tan A = 11/60

Formulas used:

tan A = Perpendicular/Base, where A is an acute angle

Pythagoras Theorem 

In triangle xyz, xz = Hypotenuse

(xz)² = (xy)² + (yz)²

Calculation:

(xz)² = 11² + 60² 

⇒ xz = √3721 = 61 

Now,  (4cos A - 7sin A) 

⇒ 4 × B/H - 7 × P/H 

⇒ 4 × 60/61 - 7 × 11/61 

⇒ 240/61 - 77/61 = 163/61 

⇒ \(2\frac{{41}}{{61}}\)

∴ (4cos A - 7sin A) = \(2\frac{{41}}{{61}}\) 

Concept:

• sin 60° = \(\frac{\sqrt{3}}{2}\)

• cos 30° = \(\frac{\sqrt{3}}{2}\)

Calculation:

Since, sin (A + B) = \(\frac{\sqrt{3}}{2}\)

⇒ sin (A + B) = sin 60° 

⇒ A + B = 60°      ----(1)

Also, cos (A - B) = \(\frac{\sqrt{3}}{2}\)

⇒ cos (A - B) = cos 30° 

⇒ A - B = 30°      ----(2)

On solving equations (1) & (2), we get,

A = 45° and B = 15° 

Hence, acute angles of A and B are 45° and 15° respectively.

Given:

We have to evalute 2 tan2 45° + cos2 30° - sin2 60°

Formula Used:

tan45° = 1 

Cos30° = √3/2

Sin60° = √3/2

Calculation:

2 tan² 45° + cos² 30° - sin² 60°

⇒ 2 × 1² + (√3/2)² – (√3/2)²

⇒ 2 + 3/4 – 3/4

⇒ 2

∴ The value of 2 tan2 45° + cos2 30° - sin2 60° is 2.