Practice Trigonometric ratios and identities - quantitative aptitude Online Quiz (set-1) For All Competitive Exams

Given, $\text"sin θ + cos θ"$ = 1

Squaring both sides,

$(sin^2 θ + cos^2 θ) + 2 \text"sin θ cos θ" = 1$

⇒ 1 + 2 sin θ cos θ = 1 ⇒ sin θ cos θ = 0.

tan x = 1 = tan 45°

∴ x = 45°

2 sin x. cos x = 2 sin (45°) . cos (45°)

= 2 × $1/{√2} × 1/{√2}$ = 1

$({sin 35°}/{cos 55°})^2 - ({cos 55°}/{sin 35°})^2$ + 2 sin 30° ....(1)

We know that

sin $({π}/2 - θ)$ = cos θ

⇒ sin (90 - 55°) = cos 55°

⇒ sin 35° = cos 55°

So from (1) we get

$({sin 35°}/{sin 35°})^2 - ({cos 55°}/{cos 55°})^2 + 2 × 1/2$

= $(1)^2 - (1)^2 + 1$

∴ Option (c) is correct.

sin 25° sin 35° sec 65° sec 55°

= sin 25° . sin 35° . $1/{cos 65°} . 1/{cos 55°}$

= sin 25° . sin 35° . $1/{cos (90 - 25°)} . 1/{cos (90 - 35°)}$

= sin 25° . sin 35° . $1/{sin 25°} . 1/{sin 35°}$ = 1

We have,

$sin^2x$ + sin x = 1...(1)

∴ sin x = 1 - $sin^2 x = cos^2 x$

On cubing equation (1), we get

$(sin^2 x + sin x)^3 = {1}^3$

$sin^6 x + sin^3 x + 3 sin^2 x. sin x (sin^2 \text"x + sin x")$ = 1

$sin^6x + sin^3 x + 3 sin^5x + 3 sin^4x$ = 1

∴ $cos^{12}x + 3 cos^{10}x + 3 cos^8 x + co^6 x$ = 1

$cos^2 \text"x + cos x = 1"$

⇒ $\text"cos x = 1" - cos^2 x = sin^2 x$

= $sin^{12}x + 3 sin^{10}x + 3 sin^8x + sin^6x$

= $sin^6x[sin^6x + 3 sin^4x + 3 sin^2 x + 1]$

= $sin^6x[sin^2 x + 1]^3$

= $[sin^4x + sin^2 x]^3$

$(∴ sin^4 x = cos^2 x)$

= $(sin^2 x + cos^2 x)$ = 1

tan1° tan2° tan3° tan4°.......tan 89°

tan1° tan2° ......tan45°......tan 89°

= 1

Given that:

$a^2 = {\text"1 + 2 sin θ cos θ"}/{\text"1 - 2 sin θ cos θ"}$

⇒ $a^2 = {(sin^2 θ + cos^2 θ) + \text"2 sin θ . cos θ"}/{(sin^2 θ + cos^2 θ) - \text"2 sin θ . cos θ"}$

⇒ $a^2 = {(\text"sin θ + cos θ")^2}/{(\text"sin θ - cos θ")^2} ⇒ a/1 = {\text"sin θ + cos θ"}/{\text"sin θ - cos θ"}$

(applying componendo dividendo formula)

⇒ ${a + 1}/{a - 1} = {(\text"sin θ + cos θ") + (\text"sin θ - cos θ")}/{(\text"sin θ + cos θ") - (\text"sin θ - cos θ")}$

⇒ ${a + 1}/{a - 1} = {\text"2 sin θ"}/{\text"2 cos θ"}$ = tan θ

Min Value = $(√{25} + 1)^2$ = 36

∵ 5 sin θ + 12 cos θ = 12

Now, squaring both sides, we get

$25 sin^2 θ + 144 cos^2 θ + \text"120 sin θ cos θ"$ = 169

⇒ 25$(1 - cos^2 θ) + 144(1 - sin^2 θ) + \text"120 sin θ cos θ"$ = 169

⇒ 25 - 25 $ cos^2 θ + 144 - 144 sin^2 θ + \text"120 sin θ cos θ"$ - 169

⇒ 25 $cos^2 θ + 144 sin^2 θ - \text"120 sin θ cos θ" = 169 - 169$

⇒ $(\text"5 cos θ - 12 sin θ")^2$ = 0

∴ 5 cos θ - 12 sin θ = 0

tan(x + 40)° tan (x + 20)° tan 3x° tan (70 - x)° tan (50 - x)° = 1

⇒ tan(x + 40) tan(x + 20) tan 3x cot [90 - (70 - x)] cot [90 - (50 - x)] = 1

⇒ tan(x + 40) tan (x + 20) tan 3x cot (x + 20) cot (x + 40) = 1

⇒ tan 3x = tan 45°

⇒ 3x = 45°

⇒ x = 15°

So, option (c) is correct.

Given,

In ΔABC,

tan B = $k/{√3 k}$

By Pythogaros theorem,

$AB^2 + AC^2 = BC^2$

⇒ $(√3 k)^2 + (1k)^2 = BC^2$

⇒ $BC^2 = 4k^2$

⇒ BC = 2k

BC = 15 cm and sin B = $4/5$

sin B = ${AC}/{AB} = 4/5$

then BC = 3m

But, BC = 15 (given)

then AC = 4 × 5 = 20

AB = 5 × 5 = 25.

Hence, the value of AB is 25 cm.

tan θ + sec θ = 2 ......(i)

As we know

⇒ $sec^2 θ - tan^2 θ$ = 1

⇒ (sec θ - tan θ)(sec θ + tan θ) = 1

⇒ sec θ - tan θ = $1/2$ ....(ii)

equation (i) - eq (ii)-

2 tan θ = 2 - 1/2

2 tan θ = $3/2 ⇒ tan θ = 3/4$

So, option (b) is correct.

$cosec^4 α - cot^4$ α = 17 (Given)

⇒ $(cosec^2 α - cot^2 α) (cosec^2 α + cot^2 α)$ = 17

⇒ 1.$({1 + cos^2 α}/{sin^2 α})$ = 17

⇒ 2 - $sin^2 α = 17 sin^2 α$

⇒ $18 sin^2 α = 2 ⇒ sin^2 α = 1/9$

∴ sin α = $1/3$ (since, α lie in first quadrant)

${\text"sec x"}/{\text"cot x + tan x"}$

⇒ ${1/{\text"cos x"}}/{{\text"cos x"}/{\text"sin x"} + {\text"sin x"}/{\text"cos x"}}$

⇒ ${1/{\text"cos x"}}/{{cos^2 x + sin^2 x}/{\text"sin x cos x"}}$

⇒ $1/{(\text"cos x")}$ × cos x × sin x

⇒ sin x

Given, x ${cosec^2 30° sec^2 45°}/{8 cos^2 45° sin^2 60°} = tan^2 60° - tan^2 30°$

⇒ ${x × (2)^2 × (√2)^2}/{8 × (1/{√2})^2 × ({√3}/2)^2} = (√3)^2 - (1/{√3})^2$

⇒ ${x × 4 × 2 × 4}/{8 × 1/2 × 3} = 3 - 1/3$

⇒ ${8x}/3 = 8/3 ⇒$ x = 1

Statement 1

1 = ${√{\text"1 - sin n"}}/{√{\text"1 + sin x"}} = √{{(\text"1 - sin x")(\text"1 - sin x")}/{(\text"1 - sin x")(\text"1 + sin x")}}$

= ${\text"1 - sin x"}/{√{1 - sin^2 x}} = {\text"1 - sin x"}/{√{cos^2 x}} = {\text"1 - sin x"}/{\text"cos x"}$

P = q

r = ${\text"cos x"}/{\text"1 + sin x"} = {\text"cos x" (\text"1 - sin x")}/{(\text"1 + sin x")(\text"1 - sin x")} = {\text"cos x" (\text"1 - sin x")}/{1 - sin^2 x}$

= ${\text"cos x" (\text"1 - sin x")}/{cos^2 x} = {\text"1 - sin x"}/{\text"cos x"}$

P = q = r

Now,

Statement 2

$p^2$ = qr

= ${\text"1 - sin x"}/{\text"cos x"} . {\text"cos x"}/{\text"1 + sin x"} = {\text"1 - sin x"}/{\text"1 + sin x"} = P^2$

So, Both are correct.

Given, A = ${π}/6 \text"and B" = {π}/3$

I. L.H.S = sin A + sin B = sin ${π}/6 + sin {π}/3$

= $1/2 + {√3}/2 = {1 + √3}/2$

R.H.S = cos A + cos B = cos ${π}/6 + {cos π}/3$

${√3}/2 + 1/2 = {√3 + 1}/2$

⇒ sin A + sin B = cos A + cos B

II. L.H.S = tan A + tan B = tan ${π}/6 + tan {π}/3$

= $1/{√3} + √3 = 4/{√3}$

R.H.S = cot A + cot B = cot ${π}/6 + cot {π}/3$

= $√3 + 1/{√3} = 4/{√3}$

⇒ tan A + tan B = cot A + cot B

Both statements are true.

Alternate Method:

A + B = ${π}/6 + {π}/3 = {π}/2$

I. sin A + sin B = sin $({π}/2 - B) + sin ({π}/2 - A)$

= cos B + cos A = cos A + cos B

II. tan A + tan B = tan $({π}/2 - B) + tan ({π}/2 - A)$

= cot B + cot A = cot A + cot B

Hence, both statements are true.

Given that, cos θ + $√3$ sin θ = 2

⇒ $1/2 cos θ + {√3}/2$ sin θ = 1

⇒ sin 30° cos θ + cos 30° sin θ = 1

⇒ sin(30° + θ) = sin 90°

30° + θ = 90°

∴ θ = 60°