Multiple Pythagorean Identities MCQ Quiz - Objective Question with Answer for Multiple Pythagorean Identities - Download Free PDF
Last updated on May 23, 2025
Latest Multiple Pythagorean Identities MCQ Objective Questions
Multiple Pythagorean Identities Question 1:
If cos A + cos2A = 1 then sin2A + sin4A is equal to:
Answer (Detailed Solution Below)
Multiple Pythagorean Identities Question 1 Detailed Solution
Given:
cos A + cos2A = 1
We need to calculate sin2A + sin4A.
Formula used:
1. sin2A + cos2A = 1
2. cos A + cos2A = 1
3. Substitute cos2A to find sin2A.
Calculations:
From the given equation:
⇒ cos A + cos2A = 1
⇒ cos2A = 1 - cos A
Now, use sin2A = 1 - cos2A:
⇒ sin2A = 1 - (1 - cos A) = cos A
For sin4A:
⇒ sin4A = (sin2A)2 = (cos A)2
Now calculate sin2A + sin4A:
⇒ sin2A + sin4A = cos A + cos2A
⇒ From the given equation, cos A + cos2A = 1
∴ sin2A + sin4A = 1.
Multiple Pythagorean Identities Question 2:
If \(\sin (x)=\frac{2}{5}\) and x is an acute angle, find the exact values of cos (4x) - cos (2x).
Answer (Detailed Solution Below)
Multiple Pythagorean Identities Question 2 Detailed Solution
Given:
sin(x) = 2/5, x is an acute angle
Formula used:
cos(4x) - cos(2x)
sin2(x) + cos2(x) = 1
cos(2x) = 2cos2(x) - 1
cos(4x) = 2cos2(2x) - 1
Calculation:
cos2(x) = 1 - sin2(x)
cos2(x) = 1 - (2/5)2
cos2(x) = 21/25
cos(x) = √(21/25) = √21 / 5
cos(2x) = 2cos2(x) - 1
cos(2x) = 2(21/25) - 1
cos(2x) = 42/25 - 25/25 = 17/25
cos(4x) = 2cos2(2x) - 1
cos(4x) = 2(17/25)2 - 1 = 578/625 - 1
cos(4x) = - 47/625
cos(4x) - cos(2x) = -47/625 - 17/25
cos(4x) - cos(2x) = -472/625
∴ cos(4x) - cos(2x) = -472/625
Multiple Pythagorean Identities Question 3:
If a sin3X + b cos3X = sinX cosX and a sinX = b cosX, then find the value of a2 + b2, provided that x is neither 0° nor 90°.
Answer (Detailed Solution Below)
Multiple Pythagorean Identities Question 3 Detailed Solution
Shortcut Trick
Let X = 45°
a sin3X + b cos3X = sinX cosX
⇒ a (1/√2)3 + b (1/√2)3 = (1/√2)(1/√2)
⇒ (1/√2)3 (a + b) = 1/2
⇒ {1/2√2} (a + b) = 1/2
⇒ (a + b) = √2 ---(1)
a sinX = b cosX
⇒ a (1/√2) = b(1/√2)
⇒ a = b
So, a = b = √2/2 = 1/√2
Now, a2 + b2= (1/√2)2 + (1/√2)2 = 1/2 + 1/2 = 1
∴ The correct answer is option (2).
Alternate Method
Given:
a sin3 X + b cos3 X = sin X cos X ---(1)
a sin X = b cos X ---(2)
Formula used:
sin2 X + cos2 X = 1
Calculations:
From equation (2):
a sin X = b cos X
⇒ a = b (cos X / sin X)
Now, from equation (1):
a sin3 X + b cos3 X = sin X cos X
⇒ b (cos X / sin X) × sin3X + b cos3 X = sin X cos X
⇒ b cos X × sin2X + b cos3 X = sin X cos X
⇒ b cos X (sin2X + cos2X) = sinX cosX
⇒ b cos X (1) = sinX cosX
⇒ b = sin X
So, a = b (cos X / sin X) = sinX (cosX / sinX) = cos X
Now,
a2 + b2
⇒ cos2X + sin2X = 1
∴ The value of a2 + b2 is 1.
Multiple Pythagorean Identities Question 4:
If x = a sec θ cos ϕ, y = b sec θ sin ϕ and z = c tan θ, then the value of \(\frac{x^2}{a^2} + \frac{y^2}{b^2} - \frac{z^2}{c^2}\) is equal to:
Answer (Detailed Solution Below)
Multiple Pythagorean Identities Question 4 Detailed Solution
Given:
x = a sec θ cos ϕ
y = b sec θ sin ϕ
z = c tan θ
Formula Used:
\(\frac{x^2}{a^2} + \frac{y^2}{b^2} - \frac{z^2}{c^2}\)
Calculation:
Substitute the values of x, y, and z:
x = a sec θ cos ϕ, y = b sec θ sin ϕ, and z = c tan θ
\(\frac{x^2}{a^2}\) = \(\frac{(a sec θ cos ϕ)^2}{a^2}\)
⇒ sec2θ cos2 ϕ
\(\frac{y^2}{b^2}\) = \(\frac{(b sec θ sin ϕ)^2}{b^2}\)⇒ sec2θ sin2 ϕ
\(\frac{z^2}{c^2}\) = \(\frac{(c tan θ)^2}{c^2}\)⇒ tan2 θ
Now solve the equation:-
⇒ \(\frac{x^2}{a^2} + \frac{y^2}{b^2} - \frac{z^2}{c^2}\) = sec2θ cos2 ϕ + sec2θ sin2 ϕ - tan2θ
⇒ sec2θ (cos2 ϕ + sin2 ϕ) - tan2θ
cos2 A + sin2 A = 1 then,
⇒ sec2θ - tan2θ
and we know, sec2 A - tan2 A = 1
So,
⇒ sec2θ - tan2θ = 1 = \(\frac{x^2}{a^2} + \frac{y^2}{b^2} - \frac{z^2}{c^2}\)
Multiple Pythagorean Identities Question 5:
\(\text{If } p = \frac{\sin A}{1 + \cos A}, \text{ then } \frac{\sin A}{1 - \cos A} \text{ is equal to:}\)
Answer (Detailed Solution Below)
Multiple Pythagorean Identities Question 5 Detailed Solution
Given:
p = \(\frac{\sin A}{1 + \cos A}\)
Formula used:
We need to find \(\frac{\sin A}{1 - \cos A}\) in terms of p.
Calculations:
We know that p = \(\frac{\sin A}{1 + \cos A}\)
⇒ Multiply both numerator and denominator of the expression \(\frac{\sin A}{1 - \cos A}\) by \(\frac{1 + \cos A}{1 + \cos A}\) to simplify:
⇒ \(\frac{\sin A}{1 - \cos A} \times \frac{1 + \cos A}{1 + \cos A} = \frac{\sin A(1 + \cos A)}{(1 - \cos A)(1 + \cos A)}\)
⇒ Using the identity \((1 - \cos^2 A) = \sin^2 A\), we get:
⇒ \(\frac{\sin A(1 + \cos A)}{\sin^2 A}\)
⇒ \(\frac{1 + \cos A}{\sin A}\)
⇒ Now, using the given value of p = \(\frac{\sin A}{1 + \cos A}\), we can write:
⇒ \(\frac{1}{p}\)
Therefore, \(\frac{\sin A}{1 - \cos A} = \frac{1}{p}\).
Top Multiple Pythagorean Identities MCQ Objective Questions
The value of tan2θ + cot2θ - sec2θ cosec2θ is:
Answer (Detailed Solution Below)
Multiple Pythagorean Identities Question 6 Detailed Solution
Download Solution PDFGiven:
tan2θ + cot2θ - sec2θ cosec2θ
Concept used:
1. tanα = sinα/cosα
2. cotα = 1/tanα
3. secα = 1/cosα
4. cosecα = 1/sinα
5. (a + b)2 - 2ab = a2 + b2
6. sin2α + cos2α = 1
Calculation:
tan2θ + cot2θ - sec2θ cosec2θ
⇒ \(\frac {sin^2θ}{cos^2θ} + \frac {cos^2θ}{sin^2θ} - \frac {1}{sin^2θ \times cos^2θ}\)
⇒ \(\frac {sin^4θ + cos^4θ - 1}{sin^2θ \times cos^2θ}\)
⇒ \(\frac {(sin^2θ + cos^2θ)^2 - 2sin^2θ cos^2θ - 1}{sin^2θ \times cos^2θ}\)
⇒ \(\frac {(1)^2 - 2sin^2θ cos^2θ - 1}{sin^2θ \times cos^2θ}\)
⇒ \(\frac {-2sin^2θ cos^2θ}{sin^2θ \times cos^2θ}\)
⇒ -2
∴ The required answer is -2.
Shortcut Trick
Use value putting method to solve this question,
Use θ = 45°
tan2θ + cot2θ - sec2θ cosec2θ
⇒ 12 + 12 - (√2)2(√2)2
⇒ 1 + 1 - 4
⇒ 2 - 4 = - 2
∴ The correct answer to this question is -2.
If a cot θ + b cosec θ = p and b cot θ + a cosec θ = q then p2 - q2 is equal to ______.
Answer (Detailed Solution Below)
Multiple Pythagorean Identities Question 7 Detailed Solution
Download Solution PDFGiven:
a cot θ + b cosec θ = p
b cot θ + a cosec θ = q
Formula used:
Cosec2 θ - cot2 θ = 1
Calculation:
a cot θ + b cosec θ = p
Squaring both sides
(a cot θ + b cosec θ)2 = (p)2
a2 cot2 θ + b2 cosec2 θ + 2 × ab cot θ × cosec θ = p2 ----- (1)
b cot θ + a cosec θ = q
Squaring both sides
(b cot θ + a cosec θ)2 = (q)2
b2 cot2 θ + a2 cosec2 θ + 2 × ab cot θ × cosec θ = q2 ----- (2)
Substracting the eq.(1) and (2)
⇒ (p2 - q2) = a2 cot2 θ + b2 cosec2 θ + 2 × ab × cot θ × cosec θ - (b2 cot2 θ + a2 cosec2 θ + 2 × ab × cot θ × cosec θ)
⇒ a2 cot2 θ - a2 cosec2 θ + b2 cosec2 θ - b2 cot2 θ
⇒ a2 (cot2 θ - cosec2 θ) + b2 (cosec2 θ - cot2 θ)
⇒ b2 - a2
If tan2 θ = 1 - a2, then the value of sec θ + tan3 θ cosec θ is:
Answer (Detailed Solution Below)
Multiple Pythagorean Identities Question 8 Detailed Solution
Download Solution PDFGiven:
tan2 θ = 1 - a2
Concept Used:
Secθ = 1/cosθ, tanθ = sinθ/cosθ, sinθ = 1/cosecθ
Sec2θ – tan2θ = 1
Calculation:
Secθ = 1/cosθ, tanθ = sinθ/cosθ, sinθ = 1/cosecθ
Sec2θ – tan2θ = 1
We have to calculate, sec θ + tan3 θ cosec θ
\(\begin{array}{l} = \frac{1}{{\cos \theta }} + \frac{{si{n^3}\theta }}{{co{s^3}\theta }} \times \frac{1}{{sin\theta }}\\ = \frac{1}{{\cos \theta }}\left[ {1 + \frac{{si{n^2}\theta }}{{co{s^2}\theta }}} \right]\\ = \frac{1}{{\cos \theta }} \times \frac{1}{{co{s^2}\theta }} = \frac{1}{{co{s^3}\theta }} \end{array}\)
Now, given that,
tan2θ = 1 – a2
⇒ sec2θ – 1 = 1 – a2
\(\begin{array}{l} \Rightarrow \frac{1}{{co{s^2}\theta }} = 2 - {a^2}\\ \Rightarrow \frac{1}{{co{s^3}\theta }} = {\left[ {2 - {a^2}} \right]^{\frac{3}{2}}} \end{array}\)
∴ Option 3 is the correct answer.
If sec2 A + tan2 A = \(\frac{4}{{17}}\), then sec4 A - tan4 A is equal to :
Answer (Detailed Solution Below)
Multiple Pythagorean Identities Question 9 Detailed Solution
Download Solution PDFGiven:
sec2 A + tan2 A = \(\frac{4}{{17}}\)
Concept Used:
sec2 A - tan2 A = 1
a2 - b2 = (a + b)(a - b)
Calculation:
⇒ sec4 A - tan4 A = (sec2 A + tan2 A )(sec2 A - tan2 A )
⇒ sec4 A - tan4 A = 4/17 × 1
⇒ sec4 A - tan4 A = 4/17
∴ Option 3 is the correct answer.
cot2A - cos2A is equal to
Answer (Detailed Solution Below)
Multiple Pythagorean Identities Question 10 Detailed Solution
Download Solution PDFGiven:
cot2A - cos2A
Concept used:
cosec2θ - cot2θ = 1
Calculation:
cot2A - cos2A
⇒ \( \rm \frac{cos^2A}{sin^2A} - cos^2A\)
⇒ \( \rm cos^2A(\frac{1}{sin^2A} - 1)\)
⇒ cos2A(cosec2A - 1)
⇒ cos2A.cot2A
∴ The required answer is cos2A. cot2A.
If cot x – tan x = 3/2, then what will be the value of cot x + tan x?
Answer (Detailed Solution Below)
Multiple Pythagorean Identities Question 11 Detailed Solution
Download Solution PDFGiven:
cot x – tan x = 3/2
Concept Used:
If x - 1/x = p, then x + 1/x = √(p2 + 4)
tan x = 1/cot x
Calculation:
cot x – tan x = 3/2
⇒ cot x – 1/cot x = 3/2
Using If x - 1/x = p, then x + 1/x = √(p2 + 4)
⇒ cot x + 1/cot x = √[(3/2)2 + 4]
⇒ cot x + 1/cot x = √(25/4)
⇒ cot x + 1/cot x = 5/2
⇒ cot x + tan x = 5/2
∴ Option 3 is the correct answer.
If \(\rm \left(\frac{\cos A}{1-\sin A}\right)+\rm \left(\frac{\cos A}{1+\sin A}\right)=4\), then what will be the value of A? (0° < A < 90°)
Answer (Detailed Solution Below)
Multiple Pythagorean Identities Question 12 Detailed Solution
Download Solution PDFCalculation :
⇒ \(\frac{cosA(1 + sinA) + cosA(1 - sinA)}{{(1 + sinA)}{(1 -sinA)}}\) = 4
⇒ \(\frac{cosA + cosA.sinA + cosA - cosAsinA}{{(1 - sinA)}{(1 + sinA)}}\) = 4
⇒ 2/cosA = 4
⇒ cosA = 1/2
⇒ A = 60°
∴ The correct answer is 60°.
If tan A + cot A = 2, then the value of 2(tan2 A + cot2 A) is :
Answer (Detailed Solution Below)
Multiple Pythagorean Identities Question 13 Detailed Solution
Download Solution PDFGiven:
tan A + cot A = 2
Calculation:
According to the question,
we have tan A + cot A = 2
Squaring both sides, we get
⇒ (tan A + cot A)2 = (2)2
⇒ tan2 A + cot2 A + 2(tan A)(cot A) = 4
⇒ tan2 A + cot2 A + 2(tan A)(1/tan A) = 4
⇒ tan2 A + cot2 A + 2 = 4
⇒ tan2 A + cot2 A = 2
Now,
2(tan2 A + cot2 A) = 2 × 2
⇒ 4
∴ The value of 2(tan2 A + cot2 A) is 4.
Shortcut Trick We can go with the value-putting method,
If we put A = 45° it satisfies the equation tan A + cot A = 2, so
2(tan2 A + cot2 A)
⇒ 2(1 + 1)
⇒ 4
If cos A + cos2 A = 1, then the value of sin4 A + sin6 A is:
Answer (Detailed Solution Below)
Multiple Pythagorean Identities Question 14 Detailed Solution
Download Solution PDFCalculation:
cos A + cos2 A = 1
⇒ cos A = 1 - cos2 A
⇒ cos A = sin2 A … (i)
sin4 A + sin6 A
⇒ sin2 A (sin2 A + sin4 A)
⇒ sin2 A (sin2 A + cos2 A) from equation (i)
⇒ sin2 A × 1
⇒ cos A from equation (i)
∴ The correct answer is cos A
If \(\sin θ + \cos θ = \frac{{\sqrt 3 - 1}}{{2\sqrt 2 }}\), then what is the value of tan θ + cot θ?
Answer (Detailed Solution Below)
Multiple Pythagorean Identities Question 15 Detailed Solution
Download Solution PDFGiven:
\(\sin θ + \cos θ = \frac{{\sqrt 3 - 1}}{{2\sqrt 2 }}\)
Concept used:
sin2θ + cos2θ = 1
Calculation:
\(\sin θ + \cos θ = \frac{{\sqrt 3 - 1}}{{2\sqrt 2 }}\)
⇒ \((\sin θ + \cos θ)^2 = \left(\frac{{\sqrt 3 - 1}}{{2\sqrt 2 }}\right)^2\)
⇒ \(\sin ^2θ + \cos ^2θ +2sinθ .cosθ = \left(\frac{{3 + 1-2\sqrt3.1}}{{8 }}\right)\)
⇒ \(1+2sinθ .cosθ = \frac{{4-2\sqrt3}}{{8 }}\)
⇒ \(2sinθ .cosθ = \frac{{4-2\sqrt3}}{{8 }}-1\)
⇒ \(2sinθ .cosθ = \frac{{4-2\sqrt3-8}}{{8 }}\)
⇒ \(sinθ .cosθ = \frac{{-(\sqrt3+2)}}{{8 }}\)
Now,
tan θ + cot θ = \(\rm \frac{sinθ}{cosθ}+\frac{cosθ}{sinθ}\)
⇒ \(\rm \frac{sin^2θ+cos^2θ}{sinθ.cosθ}\)
⇒ \(\rm \frac{1}{sinθ.cosθ}\)
⇒ \(\rm \frac{1}{\frac{{-(\sqrt3+2)}}{{8 }}}\)
⇒ \(\rm \frac{-8}{{{(\sqrt3+2)}}{}}\)
⇒ \(\rm \frac{-8(\sqrt3-2)}{{{(\sqrt3+2)(\sqrt3-2)}}{}}\)
⇒ \(\rm \frac{-8(\sqrt3-2)}{{{3-4}}{}}\)
⇒ \(\rm \frac{-8(\sqrt3-2)}{{{-1}}{}}\)
⇒ \( 8\left( {\sqrt3 - 2} \right)\)
∴ The required answer is \(8\left( {\sqrt3 - 2} \right)\).