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

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:

\[ \boxed{2} \]

Calculation

Given:

Triangle ABC is right-angled at C.

\(\tan A = \frac{BC}{AC} = \frac{a}{b}\)

\(\tan B = \frac{AC}{BC} = \frac{b}{a}\)

Therefore, \(\tan A + \tan B = \frac{a}{b} + \frac{b}{a}\)

⇒ \(\tan A + \tan B = \frac{a^2 + b^2}{ab}\)

By the Pythagorean theorem, in a right-angled triangle, \(a^2 + b^2 = c^2\), where c is the hypotenuse.

So, \(\tan A + \tan B = \frac{c^2}{ab}\)

Hence option 3 is correct

Given:

f(p) = sin^{p + 2}x + cos^{p + 2}x

Calculation:

To find the value of 6f(2) - 4f(4) + 10f(0), first calculate f(p) for p = 2, p = 4, and p = 0:

For p = 2:

f(2) = sin⁴x + cos⁴x

For p = 4:

f(4) = sin⁶x + cos⁶x

For p = 0:

f(0) = sin²x + cos²x

We know that sin²x + cos²x = 1.

Using the identity for sin⁴x + cos⁴x:

sin⁴x + cos⁴x = (sin²x + cos²x)² - 2sin²x cos²x

sin⁴x + cos⁴x = 1 - 2sin²x cos²x

Using the identity sin²x cos²x = (1/4)sin²(2x):

sin⁴x + cos⁴x = 1 - (1/2)sin²(2x)

To find sin⁶x + cos⁶x, use the identity:

sin⁶x + cos⁶x = (sin²x + cos²x)³ - 3sin²x cos²x(sin²x + cos²x)

sin⁶x + cos⁶x = 1 - 3(sin²x cos²x)

sin⁶x + cos⁶x = 1 - (3/4)sin²(2x)

Now calculate 6f(2) - 4f(4) + 10f(0):

6f(2) = 6(sin⁴x + cos⁴x) = 6(1 - (1/2)sin²(2x))

4f(4) = 4(sin⁶x + cos⁶x) = 4(1 - (3/4)sin²(2x))

10f(0) = 10(sin²x + cos²x) = 10

Combine:

6f(2) - 4f(4) + 10f(0)

= 6[1 - (1/2)sin²(2x)] - 4[1 - (3/4)sin²(2x)] + 10

= 6 - 3sin²(2x) - 4 + 3sin²(2x) + 10

= 6 - 4 + 10

= 12

The value of 6f(2) - 4f(4) + 10f(0) is 12.

Concept:

Formula: 2sin A sin B = cos(A - B) - cos(A + B)

cos c + cod d = 2cos\(c+d\over2\)cos\(c-d\over2\)

Explanation:

sin 10°·sin 50° + sin 50°·sin 250° + sin 250°·sin 10° 

= \(\frac12\) (2sin 10°·sin 50° + 2sin 50°·sin 250° + 2sin 250°·sin 10°)

= \(\frac12\)(cos 40° - cos 60° + cos 200° - cos 300° + cos 240° - cos 260°)

= \(\frac12\){cos 40° - \(\frac12\) + cos 200° - cos (360° - 60°) + cos (180° + 60°) - cos 260°} 

= \(\frac12\){cos 40° - \(\frac12\) + cos 200° - \(\frac12\) - \(\frac12\) - cos 260°} 

= \(\frac12\){- \(\frac32\) + cos 200° + cos 40° - cos 260°}

= \(\frac12\){- \(\frac32\) + 2cos120° cos80° - cos260°}

= \(\frac12\){- \(\frac32\) + 2cos(180° - 60°) cos80° - cos260°}

= \(\frac12\){- \(\frac32\) - 2cos 60° cos80° - cos260°}

= \(\frac12\){- \(\frac32\) - cos80° - cos260°}

= \(\frac12\){- \(\frac32\) - (cos80° + cos260°)}

= \(\frac12\){- \(\frac32\) - (cos80° + cos(180° + 80°))}

= \(\frac12\){- \(\frac32\) - (cos80° - cos80°)}

= \(-\frac{3}{4}\)

Option (2) is true.

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

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

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 \(-\frac{4}{5}\).

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

∴ 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.

Concept:

tan x = \(\rm \frac{Perpendicular}{Base}\) = \(\rm \frac{P}{B}\) 

sin x = \(\rm \frac{Perpendicular}{Hypotenuse}\) = \(\rm \frac{P}{H}\) 

cos x = \(\rm \frac{Base}{Hypotenuse}\) = \(\rm \frac{B}{H}\) 

By Pythagoras theorem

Hypotenuse2 = Perpendicular2 + Base2 

Calculation:

Given:

\(\rm tan\:x=-\dfrac{3}{4}\)

P= 3, B = - 4

By Pythagoras theorem

H² = P² + B²

=  3²  + (-4)²

Hypotenuse(H) = 5

sin x  = \(\rm \frac{3}{5}\)

cos x = \(\rm \frac{-4}{5}\)

sin x . cos x = \(\rm \frac{3}{5} . \rm \frac{-4}{5}\) = \(\rm \frac{-12}{25}\)