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

Given:

If p sin A - cos A = 1

Formula Used:

sin 90^∘ = 1

cos 90∘ = 0

Calculation:

p sin A - cos A = 1

To get the value of p put A = 90^∘

\( (p \sin 90^\circ - \cos 90^\circ) = 1 \)

p × 1 - 0 = 1

p = 1

Putting A = 90° and p = 1 in p2 - (1 + p2) cos A:

p2 - (1 + p2) cos A = 1² - (1 - 1²)cos 90° 

⇒ 1 - (1 + 1) × 0 

⇒ 1 - 2 × 0 = 1 - 0 = 1

∴ The value of p2 - (1 + p2) cos A = 1.

Alternate Method

Concept used:

if cosecθ + cotθ = x, then cosecθ - cotθ = 1/x

Calculation:

p sin A - cos A = 1

⇒ p sin A = 1 + cos A

⇒ p = (1 + cos A)/ sinA

⇒ p = cosec A + cot A

So, cosec A - cot A = 1/p

Now,

p + 1/p = 2 cosec A

p - 1/p = 2 cot A

So, p2 - (1 + p2) cos A

⇒ p²- p(1/p + p) cosA

⇒ p² - p(2cosecA) cosA

⇒ p²- p (2/sin A) cosA

⇒ p² - p (2 cotA)

⇒ p² - p (p - 1/p)

⇒ p² - p² + 1 = 1

∴ The correct answer is option (1).

Given: 

If sec θ + tan θ = x, find sinθ.

Formulae Used:

(secθ + tanθ) (secθ - tanθ) = 1

Solution:

sec θ + tan θ = x   ---(1)

So, sec θ - tan θ = 1/x  ---(2)

Subtracting equation (2) from equation (1):

sec θ + tan θ - (sec θ - tan θ) = x - 1/x

⇒ sec θ + tan θ - sec θ + tan θ = (x² - 1)/x

⇒ 2 tan θ = (x² - 1)/x

⇒ tan θ = (x2 - 1)/2x

We know that tan θ = p/b

So, p = (x² - 1), b = 2x

h² = p² + b² = (x² - 1)² + (2x)²

⇒ h² = (x²)² + 1 - 2x² + 4x²

⇒ h2 = (x2)2 + 1 + 2x2

⇒ h2 = (x² + 1)²

⇒ h = (x² + 1)

So, sin θ = p/h = (x2 - 1) / (x2 + 1)

∴ The correct answer is option (3).

Given:

cot² θ - 2 cos² θ = 0, (0°)

Formula Used:

cot² θ = cos² θ / sin² θ

Calculation:

Using the identity cot² θ = cos² θ / sin² θ

⇒ (cos² θ / sin² θ) - 2 cos² θ = 0

⇒ cos² θ (1 / sin² θ - 2) = 0

⇒ 1 / sin² θ - 2 = 0

⇒ 1 / sin² θ = 2

⇒ sin² θ = 1 / 2

⇒ sin θ = 1 / √2

⇒ θ = 45º

The correct answer is option 1

Given:

10sin²θ + 6cos²θ = 7

0 < θ < 90º

Formula Used:

sin²θ + cos²θ = 1

tanθ = sinθ / cosθ

Calculation:

Let sin²θ = x, then cos²θ = 1 - x

⇒ 10x + 6(1 - x) = 7

⇒ 10x + 6 - 6x = 7

⇒ 4x = 1

⇒ x = 1/4

⇒ sin²θ = 1/4

⇒ sinθ = 1/2

Since 0 < θ < 90º

⇒ sinθ = 1/2

⇒ cos²θ = 1 - (1/4)

⇒ cos²θ = 3/4

⇒ cosθ = √(3)/2

Since 0 < θ < 90º

⇒ cosθ = √(3)/2

⇒ tanθ = sinθ / cosθ

⇒ tanθ = \(\frac{(1/2)}{(√(3)/2)}\)

⇒ tanθ = 1/√(3)

The correct answer is option 3

Given:

If 3 tan A = 4 and A is an acute angle.

Formula Used:

We know that tan A = sin A / cos A

Also, sin²A + cos²A = 1

Calculation:

3 tan A = 4

⇒ tan A = 4/3

⇒ sin A / cos A = 4/3

Let sin A = 4k and cos A = 3k

sin²A + cos²A = 1

⇒ (4k)² + (3k)² = 1

⇒ 16k² + 9k² = 1

⇒ 25k² = 1

⇒ k² = 1/25

⇒ k = 1/5

So, sin A = 4k = 4/5

and cos A = 3k = 3/5

Now, 4 sin A + 3 cos A = 4(4/5) + 3(3/5)

⇒ 16/5 + 9/5

⇒ 25/5

⇒ 5

The value of 4sinA + 3cosA is 5.

Given:

sec θ + tan θ = 5

Concept used:

If sec θ + tan θ = y

then sec θ - tan θ = 1/y

Calculation:

sec θ + tan θ = 5  ----- (1)

then,

sec θ - tan θ = 1/5 ------- (2)

Subtracting the eq. (1) and (2)

⇒ (sec θ + tan θ) - (sec θ - tan θ) = (5 - 1/5)

⇒ sec θ + tan θ - sec θ + tan θ = 24/5

⇒ 2 × tan θ = 24/5

⇒ tan θ = 12/5

∴ The correct answer is 12/5.

Given:

{(3Sinθ – Cosθ)/(Cosθ + Sinθ)} = 1

Calculation:

We have a trigonometric equation 

{(3Sinθ – Cosθ)/(Cosθ + Sinθ)} = 1

On dividing numerator and denominator by sin θ, we get

⇒ [{(3sinθ – cosθ)/Sinθ}/{(cosθ + sinθ)/sinθ}] = 1

⇒ {(3 – cotθ)/(cotθ + 1)} = 1

⇒ 3 – cotθ = 1 + cotθ

⇒ 2cotθ = 2

cotθ = 1

The value is 1.

Given:

sec2 θ + tan2 θ = 25/18

Formula used:

sec2 θ - tan2 θ = 1

(sec2 θ + tan2 θ)(sec2 θ - tan2 θ) = sec4 θ - tan4 θ

Calculation

⇒ (sec2 θ + tan2 θ)(sec2 θ - tan2 θ) = sec4 θ - tan4 θ

⇒ 25/18 × 1 = sec4 θ - tan4 θ

⇒ sec4 θ - tan4 θ = 25/18

The value is 25/18.

Given:

sec A + tan A = 5

Concept used:

If sec A + tan A = x then,

⇒ sec A - tan A = 1/x

Formula used:

Sin θ = P/H ; sec θ = H/B ; tan θ = P/B

Pythagorean theorem:

 H2 = P2 + B2

Where, H = hypotenuse ; P = perpendicular ; B = base

Calculation:

⇒ sec A + tan A = 5 ------- (1)

then,

⇒ sec A - tan A = 1/5 -------- (2)

Adding the eq.(1) and (2)

⇒ Sec A + tan A + sec A - tan A = 5 + (1/5)

⇒ 2 × sec A = 26/5

⇒ sec A = 13/5 = H/B

By Pythagorean theorem:

⇒ H² = P² + B²

⇒ (13)² = P² + 5²

⇒ P² = 169 - 25 = 144

⇒ P = √144 = 12 

Sin A = P/H = 12/13

∴ The correct answer is 12/13.

Calculation

sin t + cos t = \(\frac{4}{5}\)

Squaring both sides,

(sin t + cos t)² = (4/5)²

​⇒ (sin²t + cos²t + 2 × sin t × cos t  = 16/25

⇒ 1 + 2 × sin t × cos t = 16/25 ......(sin2t + cos2t  = 1)

⇒ 2 × sin t × cos t = 16/25 - 1

⇒ 2 × sin t × cos t = -9/25

⇒ sin t × cos t = -9/50

Given

\(\frac{\sin^3 α + \cos^3 α}{\sin α + \cos α}\)

Formula Used:

a³ + b³ = (a + b) (a² + b² - ab)

sin²θ + cos²θ = 1

Calculation:

\(\frac{\sin^3 α + \cos^3 α}{\sin α + \cos α}\) = [(sin α + cos α) (sin²α + cos²α - sinα cosα)]/(sin α + cos α)

⇒ 1 - sin α cos α

The value is 1 - sin α cos α.

Given

 tan2α = \(\rm\frac{2 tan \alpha}{1-tan^{2}\alpha}\)

Formula used:

Tan 15° = Tan(45 – 30)°

By the trigonometry formula, we know,

Tan (A – B) = (Tan A – Tan B) /(1 + Tan A Tan B)

Calculation

Therefore, we can write,

tan(45 – 30)° = tan 45° – tan 30°/1+ tan 45° tan 30°

Now putting the values of tan 45° and tan 30° from the table we get;

tan(45 – 30)° = (1 – 1/√3)/ (1 + 1.1/√3)

tan (15°) = √3 – 1/ √3 + 1

= (√3 – 1)²/ [(√3)² - 1²]

= (3 + 1 - 2√3)/2 = 2 - √3

= 0.268

Hence, the value of tan (15°) is 0.268.
Alternate Method 
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tan 30° = tan 2(15°)

By the trigonometry formula, we know,

 tan2α = \(\rm\frac{2 tan \alpha}{1-tan^{2}\alpha}\),

Calculation

Therefore, we can write,

tan 30° = 2 × tan 15° /(1 - tan2 15°)

Now putting the values of tan 30° we get;

⇒ 1/ √3 = 2 tan 15° / (1 - tan2 15°)

let tan (15°) = x

⇒ 1/ √3 = 2x / (1 - x2)

⇒ x2 - 1 + 2√3 x = 0

⇒ x2 + 2√3 x - 1 = 0

From quadratic formula,

x = \(\frac{-2√3 \pm \sqrt{(2√3)^2 -4(1)(-1) }}{2\times 1}\)

⇒ x = \(\frac{-2√3 \pm \sqrt{12 +4 }}{2\times 1}\)

⇒ x = \(\frac{-2√3 \pm \sqrt{16 }}{2\times 1}\)

⇒ (4 - 2√3)/2 = 2 - √3

= 0.268

Hence, the value of tan (15°) is 0.268.

Formula Used : 

Sin²x + Cos²x = 1

Calculation : 

(cos x - sin x)2 + (cos x + sin x)2

⇒ cos²x + sin²x - 2cosx.sinx + cos²x + sin²x + 2sinx.cosx

⇒ 2cos²x + 2sin²x

⇒ 2(cos²x + sin²x)

⇒ 2

∴ The correct answer is 2.

Given:

\(\rm \left(\frac{1}{\sin θ}+\frac{1}{\tan θ}\right)\rm \left(\frac{1}{\sin θ}-\frac{1}{\tan θ}\right)\)

Formula used:

(A² - B²) = (A + B) × (A - B)

Cosec² θ - cot² θ = 1

Calculation:

\(\rm \left(\frac{1}{\sin θ}+\frac{1}{\tan θ}\right)\rm \left(\frac{1}{\sin θ}-\frac{1}{\tan θ}\right)\)

⇒ (cosec θ + cot θ) × (cosec θ - cot θ)

⇒ Cosec2 θ - cot2 θ

⇒ 1

∴ The correct answer is 1. 

Concept used:

Cosθ/ Sinθ = Cotθ

Calculation:

\(\rm \frac{21\ cosA+3\ sinA}{3\ cosA+4\ sinA}\) = 2

By dividing by Sin A we get

⇒ (21 cot A + 3) / (3 cot A + 4) = 2

⇒ 15 cot A = 5

cot A = 1/3

∴ The correct option is 3