TRIGONOMETRY PRACTICE QUESTIONS: 11-20

Question 11.) If `theta` ∈ [`\frac{π}{2}`, π], and `\phi` ∈ [`\frac{3π}{2}`, 2π], then the value of which of the following will always be negative?

(A) `sin` (`\phi` - `theta`)

(B) `cos` (`\phi` - `theta`)

(C) `sin` (`\phi` + `theta`)

(D) `cos` (`\phi` + `theta`)

Answer. (B)

Solution.

Since it is given that,

`theta` ∈ [`\frac{π}{2}`, π], and `\phi` ∈ [`\frac{3π}{2}`, 2π]

Since the value of `sin``theta` in this range will always be positive, and that of `cos``theta` will always be negative. While the value of `sin``\phi` will always be negative, and that of `cos``\phi` will always be positive in the given range.

Now we will check each case given in the options by using their formula,

`sin`(`\phi` - `theta`) = `sin``\phi``cos``theta` - `cos``\phi``sin``theta`

Now the value of `sin``\phi` and `cos``theta` will always be negative, and as we know, the product of two negative numbers is always positive. So, its product will always be positive. While the value of `cos``\phi` and `sin``theta` will always be positive, the product of two positive numbers is always positive. The difference between two positive numbers may be positive or negative, depending on their magnitudes. Hence, the value of `sin`(`\phi` - `theta`) will not always be negative.

`cos`(`\phi` - `theta`) = `cos``\phi``cos``theta` + `sin``\phi``sin``theta`

Here again, the value of `cos``\phi` will always be positive while that of `cos``theta` will always be negative, and as we know, the product of a negative number and a positive number is always negative. So, its product will always be negative. Similarly, the value of `sin``\phi` will always be negative while that of `sin``theta` will always be positive, and the product of a negative number and a positive number is always negative. So, its product will always be negative. The sum of two negative numbers is always negative. Hence, the value of `cos`(`\phi` - `theta`) will always be negative.

`sin`(`\phi` + `theta`) = `sin``\phi``cos``theta` + `cos``\phi``sin``theta`

Here again, the value of `sin``\phi` and `cos``theta` will always be negative, and as we know, the product of two negative numbers is always positive. So, its product will always be positive. While the value of `cos``\phi` and `sin``theta` will always be positive, the product of two positive numbers is always positive. The sum of two positive numbers is always positive. Hence, the value of `sin`(`\phi` + `theta`) will always be positive.

`cos`(`\phi` + `theta`) = `cos``\phi``cos``theta` - `sin``\phi``sin``theta`

Here again, the value of `cos``\phi` will always be positive while that of `cos``theta` will always be negative, and as we know, the product of a negative number and a positive number is always negative. So, its product will always be negative. Similarly, the value of `sin``\phi` will always be negative while that of `sin``theta` will always be positive, and the product of a negative number and a positive number is always negative. So, its product will always be negative. But the difference between two negative numbers may be positive or negative, depending on their magnitudes. Hence, the value of `sin`(`\phi` - `theta`) will not always be negative.

Hence, the value of `cos`(`\phi` - `theta`) will always be negative.

Question 12.) If `cot` `frac{y}{2}` = ±`\sqrt{5}` `cot` `frac{x}{2}`, then find the value of 3(`cos` y - `cos` x) + 2`cos` x `cos` y.

(A) 1

(B) -1

(C) 2

(D) -2

Answer. (C)

Solution.

Given that,  `cot``frac{y}{2}` = ±`\sqrt{5}` `cot``frac{x}{2}`

Now squaring both sides, we get,

`\cot^{2}``frac{y}{2}` = 5 `\cot^{2}``frac{x}{2}`

or,       `\frac{\cos^{2}\frac{y}{2}}{\sin^{2}\frac{y}{2}}` = 5 `\frac{\cos^{2}\frac{x}{2}}{\sin^{2}\frac{x}{2}}`

or,       `\frac{2\cos^{2}\frac{y}{2}}{2\sin^{2}\frac{y}{2}}` = 5 `\frac{2\cos^{2}\frac{x}{2}}{2\sin^{2}\frac{x}{2}}`

or,       `\frac{1+\cos y}{1-\cos y}` = 5 `\frac{1+\cos x}{1-\cos x}`

or,       `\frac{1+\cos y}{1-\cos y}` = `\frac{5+5\cos x}{1-\cos x}`

Now we will apply the componendo and dividendo rule,

`cos` y = `\frac{2+3\cos x}{3+2\cos x}`

3`cos` y + 2 `cos` x `cos `y = 2 + 3 `cos` x

 or,       3(`cos` y - `cos` x) + 2`cos` x `cos` y = 2

Question 13.) If `theta` = `sin^{-1}`(`\frac{5}{\sqrt{106}}`) and 𝞪 = `cos^{-1}`(`\frac{4}{\sqrt{65}}`) then find the value `tan`(`theta` + 𝞪).

(A) 83

(B) 41

(C) 36

(D) 69

Answer. (A)

Solution.

Since it is given that,

`theta` = `sin^{-1}`(`\frac{5}{\sqrt{106}}`) and 𝞪 = `cos^{-1}`(`\frac{4}{\sqrt{65}}`)

It can be written as,

`sin` `theta` = `\frac{5}{\sqrt{106}}`

Hence,   `tan` `theta` = `\frac{5}{9}`

Similarly if, 𝞪 = `cos^{-1}`(`\frac{4}{\sqrt{65}}`)

Then, `cos`𝞪 = `\frac{4}{\sqrt{65}}`

and, `tan`𝞪 = `\frac{7}{4}`

On solving it, we get, `tan`(`theta` + 𝞪) = 83

Question 14.) If `cot` x = `\frac{\sqrt{7}}{3\sqrt{2}}` then find the value of `cos` 3x(`sin` 5x - `cos` 5x) - `sin` 3x(`sin` 5x + `cos` 5x).

(A) `\frac{2\sqrt{7}-11}{25}`

(B) `\frac{6\sqrt{7}-11}{25}`

(C) `\frac{6\sqrt{14}+11}{25}`

(D) `\frac{2\sqrt{14}+11}{25}`

Answer. (C)

Solution.

Since it is given that,

`cot` x = `\frac{\sqrt{7}}{3\sqrt{2}}`

Hence, `cos` x = `\frac{\sqrt{7}}{5}`

or,  `\cos^{2}` x = `\frac{7}{25}`

or,  2`\cos^{2}` x = `\frac{14}{25}`

or,  2`\cos^{2}` x - 1 = `\frac{14}{25}` - 1

or,  `cos` 2x = `\frac{-11}{25}`    .......(1)

and, `sin` 2x = `\frac{6\sqrt{14}}{25}`    .......(2)

Now we will evaluate the value of, `cos` 3x(`sin` 5x - `cos` 5x) - `sin` 3x(`sin` 5x + `cos` 5x)

`cos` 3x(`sin` 5x - `cos` 5x) - `sin` 3x(`sin` 5x + `cos` 5x) = `cos` 3x `sin` 5x - `cos` 3x `cos` 5x - `sin` 3x `sin` 5x - `sin` 3x `cos` 5x

= (`cos` 3x `sin` 5x - `sin` 3x `cos` 5x) - (`cos` 3x `cos` 5x + `sin` 3x `sin` 5x)

= `sin` (5x - 3x) - `cos` (5x - 3x) = `sin` 2x - `cos` 2x

Now we will put the values of `cos` 2x and `sin` 2x from equations (1) and (2)

`sin` 2x - `cos` 2x = `\frac{6\sqrt{14}}{25}` - `\frac{-11}{25}` = `\frac{6\sqrt{14}+11}{25}`

Question 15.) Find the value of `sin^{2}`20° + `sin^{2}`40° + `\sin`20°`\sin`40°.

(A) `\frac{1}{\sqrt{3}}`

(B) `\frac{3}{4}`

(C) `\frac{\sqrt{3}}{2}`

(D) 1

Answer. (B)

Solution.

The given expression is,

`sin^{2}`20° + `sin^{2}`40° + `\sin`20°`\sin`40°

or, `\frac{1}{2}` (2`sin^{2}`20° + 2`sin^{2}`40° + 2`\sin`20°`\sin`40°)

or, `\frac{1}{2}` (2`sin^{2}`20° + 2`sin^{2}`40° - (-2`\sin`20°`\sin`40°))

or, `\frac{1}{2}` (1 - `\cos`40° + 1 - `\cos`80° - `\cos`60° + `\cos`20°)

Now putting `\cos`60° = `\frac{1}{2}` we get,

`\frac{1}{2}` (1 - `\cos`40° + 1 - `\cos`80° - `\frac{1}{2}` + `\cos`20°)

= `\frac{1}{2}` (1 + 1 - `\frac{1}{2}` - (`\cos`40° + `\cos`80°) + `\cos`20°)

= `\frac{1}{2}` (`\frac{3}{2}` - 2`\cos`60°`\cos`20° + `\cos`20°)

= `\frac{1}{2}` (`\frac{3}{2}` - `\cos`20° + `\cos`20°)

= `\frac{3}{4}`

Question 16.) Two towers of different heights are made at a distance of 10`\sqrt{3}` meters from each other. If a straight rope is tied from the top of both towers, making an angle of 30° with the ground, then find the difference between the heights of the towers.

(A) `\frac{10}{\sqrt{3}}`

(B) 5

(C) 5`\sqrt{3}`

(D) 10

Answer. (D)

Solution.

In the given diagram, AB and CD are the two towers.

Let h1 and h2 be the heights of the towers AB and CD.

Now, it is given that the measure of BD = 10`\sqrt{3}` meters and the measure of ∠ AFB = 30°.

Now, as we can see in the diagram, FB  CE, So, ∠ACE = ∠ AFB = 30°, CD = BE, and CE = BD.

Now in △ ACE, `\frac{AE}{CE}` = `tan`∠ACE.

Now, AE = AB - BE = AB - CD = h1-h2.

So, `\frac{AE}{CE}` = `\frac{h1-h2}{10\sqrt{3}}` = `tan`∠ACE = `tan`30°.

or, `\frac{h1-h2}{10\sqrt{3}}` = `\frac{1}{\sqrt{3}}`.

or, h1-h2 = 10.

Question 17.) If R1 and R2 are the roots of the following equation, 

`x^{2}``\sec\theta` + `x`(`\tan\theta\sec\theta` - 1) - `\tan\theta` = 0

Among R1 and R2, R1 is positive while R2 is negative. Then, which of the following relationships between R1 and R2 will be equal to 1?

(A) `\R1^{2}` + `R2^{2}`

(B) `\R1^{2}` - `R2^{2}`

(C) `\frac{1}{\R1^{2}}` - `\frac{1}{\R2^{2}}`

(D) `\frac{1}{\R1^{2}}` + `\frac{1}{\R2^{2}}`

Answer. (C)

Solution.

The given equation is,

`x^{2}``\sec\theta` + `x`(`\tan\theta\sec\theta` - 1) - `\tan\theta` = 0    .....(1)

Suppose we have to find roots of a quadratic equation, a`x^{2}` + b`x` + c = 0.

Then we will use the following relationship to find the roots,

`\frac{-b\pm\sqrt{b^2-4ac}}{2a}`

Now, in the given equation(1),

a = `\sec\theta`, b = `\tan\theta\sec\theta` - 1, and c = -`\tan\theta`

Now, putting these values in the given relationship and further solving, we get the roots,

R1 = `\cos\theta` and R2 = -`\cot\theta`

Now we will use the following trigonometric identity,

`\sec^{2}\theta` - `\tan^{2}\theta` = 1

`\sec^{2}\theta` - `(-\tan\theta)^{2}` = `(\frac{1}{\cos\theta})^{2}` - `(\frac{1}{-\cot\theta})^{2}` = `\frac{1}{\R1^{2}}` - `\frac{1}{\R2^{2}}`

Question 18.) A man standing on top of a building of height 100 meters. He looks at the road and finds that a car is standing in the west direction, making an angle of elevation of `sin^{-1}`(`\frac{1}{3}`) from the top of the building. Suddenly, the car started moving east and stopped at a distance where the angle of elevation from the top of the building was `cos^{-1}`(`\frac{1}{3}`). Find the distance travelled by the car.

(A) 125`\sqrt{2}`

(B) 200`\sqrt{2}`

(C) 175`\sqrt{2}`

(D) 225`\sqrt{2}`

Answer. (D)

Solution.

As per the information given in the question, the following diagram represents everything.

In the diagram above, AD is the building, and B is the point west of the building from which the car started moving toward the east at point C.

Now, the measurements are, AD = 100 meters, ∠ ABD = `sin^{-1}`(`\frac{1}{3}`), and ∠ ACD = `cos^{-1}`(`\frac{1}{3}`)

We need to find the distance BC.

Now in △ ABD, `tan`∠ABD = `\frac{AD}{BD}`

or, `tan`[`sin^{-1}`(`\frac{1}{3}`)] = `\frac{100}{BD}`

or, `tan`[`tan^{-1}`(`\frac{1}{2\sqrt{2}}`)] = `\frac{100}{BD}`

or, `\frac{1}{2\sqrt{2}}` = `\frac{100}{BD}`

or, BD = 200`\sqrt{2}`

Similarly in △ ACD, `tan`∠ACD = `\frac{AD}{CD}`

or, `tan`[`cos^{-1}`(`\frac{1}{3}`)] = `\frac{100}{CD}`

or, `tan`[`tan^{-1}`(2`\sqrt{2}`)] = `\frac{100}{CD}`

or, 2`\sqrt{2}` = `\frac{100}{CD}`

or, CD = `\frac{100}{2\sqrt{2}}` = 25`\sqrt{2}`

Now, BC = BD + CD = 200`\sqrt{2}` + 25`\sqrt{2}` = 225`\sqrt{2}`

Hence, the distance travelled by the car is 225`\sqrt{2}` meters.

Question 19.) If `sin^{-1}``x`, `sin^{-1}``y`, and `sin^{-1}``z` are the measures of the angles of a triangle. Then find the value of,

2(`x``y``\sqrt{1-z^{2}}` + `y``z``\sqrt{1-x^{2}}` + `z``x``\sqrt{1-y^{2}}`).

(A) `x^{2}` + `y^{2}` + `z^{2}` + 2`x``y``z`

(B) `x^{2}` + `y^{2}` + `z^{2}`

(C) `x^{2}` + `y^{2}` + `z^{2}` + `x``y``z`

(D) `x^{2}` + `y^{2}` + `z^{2}` - `x``y``z`

Answer. (B)

Solution.

Since it is given that, `sin^{-1}``x`, `sin^{-1}``y`, and `sin^{-1}``z` are the measures of the angles of a triangle.

So, `sin^{-1}``x` + `sin^{-1}``y` + `sin^{-1}``z` = 𝛑     .....(1)

or, `sin^{-1}``x` + `sin^{-1}``y` = 𝛑 - `sin^{-1}``z`

Now taking the cosine ratio on both sides, we get,

`\cos`(`sin^{-1}``x` + `sin^{-1}``y`) = `\cos`(𝛑 - `sin^{-1}``z`)

or, `\cos``sin^{-1}``x``\cos``sin^{-1}``y` + `\sin``sin^{-1}``x``\sin``sin^{-1}``y` = -`\cos``sin^{-1}``z`

or, (`\cos``cos^{-1}``\sqrt{1-x^{2}}`)(`\cos``cos^{-1}``\sqrt{1-y^{2}}`) + (`\sin``sin^{-1}``x`)(`\sin``sin^{-1}``y`) = -`\cos``cos^{-1}``\sqrt{1-z^{2}}`

or, (`\sqrt{1-x^{2}}`) (`\sqrt{1-y^{2}}`) - `x``y` = -`\sqrt{1-z^{2}}`

or, (`\sqrt{1-x^{2}}`) (`\sqrt{1-y^{2}}`) = `x``y` -`\sqrt{1-z^{2}}`

Now, squaring both sides and further solving, we get,

2`x``y``\sqrt{1-z^{2}}` = `x^{2}` + `y^{2}` - `z^{2}`     .....(2)

Now, in the same manner,

2`y``z``\sqrt{1-x^{2}}` = -`x^{2}` + `y^{2}` + `z^{2}`     .....(3)

and, 2`z``x``\sqrt{1-y^{2}}` = `x^{2}` - `y^{2}` + `z^{2}`     .....(4)

Now adding equations (1), (2), and (3), we get

2(`x``y``\sqrt{1-z^{2}}` + `y``z``\sqrt{1-x^{2}}` + `z``x``\sqrt{1-y^{2}}`) = `x^{2}` + `y^{2}` + `z^{2}`

Question 20.) If a`\Cos``\theta` = b`Cos`(`\theta` + `\frac{2\pi}{3}`) = c`Cos`(`\theta` + `\frac{4\pi}{3}`). Then find the value of ab+bc+ca.

(A) 0

(B) `\frac{2}{3}`

(C) `\frac{7}{5}`

(D) 1

Answer. (A)

Solution.

Since it is given in the question that,

a`\Cos``\theta` = b`Cos`(`\theta` + `\frac{2\pi}{3}`) = c`Cos`(`\theta` + `\frac{4\pi}{3}`)

Let, a`\Cos``\theta` = b`Cos`(`\theta` + `\frac{2\pi}{3}`) = c`Cos`(`\theta` + `\frac{4\pi}{3}`) = k

So, now,

a = `\frac{k}{Cos(\theta)}`, b = `\frac{k}{Cos(\theta + \frac{2\pi}{3})}`, and c = `\frac{k}{Cos(\theta + \frac{4\pi}{3})}`

Now we will calculate ab+bc+ca,

Hence, ab+bc+ca = 0.