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

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.

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

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.​

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

⇒ sec2θ sin2 ϕ

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 = \(\frac{x^2}{a^2} + \frac{y^2}{b^2} - \frac{z^2}{c^2}\)

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, \(\frac{\sin A}{1 - \cos A} = \frac{1}{p}\).

Given:

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.

Given:

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 

Given:

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.

Given:

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.

Given:

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 cos​2A. cot​2A.

Given:

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.

Calculation :

⇒ \(\frac{cosA(1 + sinA) + cosA(1 - sinA)}{{(1 + sinA)}{(1 -sinA)}}\) = 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°.

Given:

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

Calculation:

 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

Given:

\(\sin θ + \cos θ = \frac{{\sqrt 3 - 1}}{{2\sqrt 2 }}\)

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 \(8\left( {\sqrt3 - 2} \right)\).