TRIGONOMETRY PRACTICE QUESTIONS: 1-10

Question 1.) If `theta` ∈ [0, 2π], then find the correct ranges of `theta` that satisfy the following equations,

`sec` `theta` - `tan` `theta` ＞ `sec` `theta` + `tan` `theta`

&, `sin` `theta` + `cos` `theta` ＞ 0

(A) `theta` ∈ (0, π) & `theta` ∈ (`\frac{3π}{2}`, 2π)

(B) `theta` ∈ (`\frac{π}{2}`, `\frac{2π}{3}`) & `theta` ∈ (π, `\frac{4π}{3}`)

(C) `theta` ∈ (`\frac{5π}{6}`, π) & `theta` ∈ (`\frac{5π}{3}`, 2π)

(D) `theta` ∈ (`\frac{3π}{4}`, π) & `theta` ∈ (`\frac{7π}{4}`, 2π)

Answer. (D)

Solution.

Since it is given that,

`sec` `theta` - `tan` `theta` ＞ `sec` `theta` + `tan` `theta`

&, `sin` `theta` + `cos` `theta` ＞ 0

Let's take the first equation,

`sec` `theta` - `tan` `theta` ＞ `sec` `theta` + `tan` `theta`

- `tan` `theta` ＞ `tan` `theta`     ⇒ `tan` `theta` < 0

Now let's take the second equation,

`sin` `theta` + `cos` `theta` ＞ 0

or, `sin` `theta` ＞ -`cos` `theta`     ⇒ `tan` `theta` ＞ -1

Hence, we got that, 0 ＞ `tan` `theta` ＞ -1

It gives the following ranges of `theta`,

`theta` ∈ (`\frac{3π}{4}`, π) & `theta` ∈ (`\frac{7π}{4}`, 2π)

Question 2.) If `cos` (`theta` - `\phi`) = `frac{5}{13}` and `sin` (`theta` + `\phi`) = `frac{4}{5}`, then find the value of `tan` `theta`.

(A) `frac{5}{6}`

(B) `frac{7}{4}`

(C) `frac{5}{2}`

(D) `frac{9}{4}`

Answer. (B)

Solution.

Since we have to find the value of `tan` `theta`. So, we will use the following method,

`tan` `theta` = `tan` (`theta/2` + `theta/2`)

= `tan` (`theta/2` + `ф/2` + `theta/2` - `ф/2`)

= `tan` ( `\frac{\theta+\phi}{2}` + `\frac{\theta-\phi}{2}`)

Now, here we will use the following relationship,

`tan` (A+B) = `\frac{\tanA+\tanB}{1-\tanA\tanB}`

But before that, we will find the values of

`tan` (`\frac{\theta+\phi}{2}`) and `tan` (`\frac{\theta-\phi}{2}`)

Since two equations have been given to us, which are as follows

`cos` (`theta` - `\phi`) = `frac{5}{13}`

and, `sin` (`theta` + `\phi`) = `frac{4}{5}`

Let's consider the first equation,

`cos` (`theta` - `\phi`) = `frac{5}{13}`

or, 2`\cos^{2}`(`\frac{\theta-\phi}{2}`) - 1 = `frac{5}{13}`  [using the formula of `cos` 2A]

On solving it we get, `\cos`(`\frac{\theta-\phi}{2}`) = `\frac{3}{\sqrt{13}}`

And, `\sin`(`\frac{\theta-\phi}{2}`) = `\frac{2}{\sqrt{13}}`

Hence, `\tan`(`\frac{\theta-\phi}{2}`) = `frac{2}{3}`

Now the second given equation is,

`sin` (`theta` + `\phi`) = `frac{4}{5}`

Now, `cos` (`theta` + `\phi`) = `frac{3}{5}`

or, 2`\cos^{2}`(`\frac{\theta+\phi}{2}`) - 1 = `frac{3}{5}`

On solving it, we will get, `\cos`(`\frac{\theta+\phi}{2}`) = `\frac{2}{\sqrt{5}}`

And, `\sin`(`\frac{\theta+\phi}{2}`) = `\frac{1}{\sqrt{5}}`

`\tan`(`\frac{\theta+\phi}{2}`) = `frac{1}{2}`

`tan` `theta` = `\frac{\frac{2}{3}+\frac{1}{2}}{1-\frac{1}{3}}` = `frac{7}{4}`

Question 3.) If `theta` = `tan^{-1}`(`\frac{1}{2}`) + `tan^{-1}`(`\frac{1}{8}`) + `tan^{-1}`(`\frac{1}{18}`) + `tan^{-1}`(`\frac{1}{32}`) + `tan^{-1}`(`\frac{1}{50}`) + `tan^{-1}`(`\frac{1}{72}`) then find the value of `sec^{2}``theta`.

(A) `frac{85}{49}`

(B) `frac{64}{25}`

(C) `frac{59}{36}`

(D) `frac{71}{16}`

Answer. (A)

Solution.

Since it is given that,

`theta` = `tan^{-1}`(`\frac{1}{2}`) + `tan^{-1}`(`\frac{1}{8}`) + `tan^{-1}`(`\frac{1}{18}`) + `tan^{-1}`(`\frac{1}{32}`) + `tan^{-1}`(`\frac{1}{50}`) + `tan^{-1}`(`\frac{1}{72}`)

It can be written as,

`theta` = `tan^{-1}`(`\frac{2}{4}`) + `tan^{-1}`(`\frac{2}{16}`) + `tan^{-1}`(`\frac{2}{36}`) + `tan^{-1}`(`\frac{2}{64}`) + `tan^{-1}`(`\frac{2}{100}`) + `tan^{-1}`(`\frac{2}{144}`)

or, `theta` = `tan^{-1}`(`\frac{3-1}{1+(3)(1)}`) + `tan^{-1}`(`\frac{5-3}{1+(5)(3)}`) + `tan^{-1}`(`\frac{7-5}{1+(7)(5)}`) + `tan^{-1}`(`\frac{9-7}{1+(9)(7)}`) + `tan^{-1}`(`\frac{11-9}{1+(11)(9)}`) + `tan^{-1}`(`\frac{13-11}{1+(13)(11)}`)

Now, here we will use the following relationship,

`tan^{-1}`(`\frac{A - B}{1 + AB }`) = `tan^{-1}`A - `tan^{-1}`B

So, `theta` = `tan^{-1}`(3) - `tan^{-1}`(1) + `tan^{-1}`(5) - `tan^{-1}`(3) + `tan^{-1}`(7) - `tan^{-1}`(5) + `tan^{-1}`(9) - `tan^{-1}`(7) + `tan^{-1}`(11) - `tan^{-1}`(9) + `tan^{-1}`(13) - `tan^{-1}`(11)

or, `theta` = `tan^{-1}`(13) - `tan^{-1}`(1) = `tan^{-1}`(`\frac{13-1}{1+(13)(1)}`)

or, `theta` = `tan^{-1}`(`frac{12}{14}`) = `tan^{-1}`(`frac{6}{7}`)

or, `tan``theta` = `frac{6}{7}`

Now we will use the following identity of trigonometry

`sec^{2}``theta` = 1 + `tan^{2}``theta`

Now putting the value of `tan``theta` and further calculation we will get,

`sec^{2}``theta` = `frac{85}{49}`

Question 4.) If ɸ, `theta` ∈ [0, π] and they satisfy the following three equations,

`tan` `theta` ＜ 0

`sin` `theta` = `frac{\sqrt{3}}{2}`

2`cos` `theta` (1 - sinɸ) = `sin^{2}``theta` (tan`theta/2` + `cot``theta/2`) cos ɸ - 1

Then find the value of ɸ.

(A) `\frac{π}{2}`

(B) `\frac{π}{4}`

(C) `\frac{2π}{3}`

(D) `\frac{π}{3}`

Answer. (D)

Solution.

Since it is given that,

2`cos` `theta` (1 - sinɸ) = `sin^{2}``theta` (tan`theta/2` + `cot``theta/2`) cos ɸ - 1

Now we will solve this equation in parts,

Hence we will get,

Hence, the main equation will be now,

2`cos` `theta` - 2`cos` `theta` sin ɸ = 2`sin` `theta` cos ɸ - 1

2`cos` `theta` = 2(`cos` `theta` `sin` ɸ + 2`sin` `theta` cosɸ) - 1

2`cos``theta` = 2`sin`(`theta`+ɸ) - 1

Now the other two equations are,

`tan``theta` ＜ 0

`sin``theta` = `frac{\sqrt{3}}{2}`

If we consider these two equations then we will find that the value of `tan``theta` is negative and the value of `sin``theta` is positive then it means that,

`frac{π}{2}` ＜ `theta` ＜ π

So,                                                         `theta` = `frac{2π}{3}`

Hence,                                   cos`theta` = cos`frac{2π}{3}` = `frac{-1}{2}`

Now putting this value in the main equation, we get,

2(`frac{-1}{2}`) = 2sin(`theta`+ɸ) - 1

-1 = 2sin(`theta`+ɸ) - 1

2sin(`theta`+ɸ) = 0

`theta`+ɸ = π

`frac{2π}{3}` + ɸ = π

ɸ = `frac{π}{3}`

Question 5.) If `\sin x` + `\sin y` + `\sin z` = 0

and, `\cos x` + `\cos y` + `\cos z` = 0

then find the value of `\tan`(`\x`-`\y`).

(A) -`\sqrt{3}`

(B) `\frac{1}{\sqrt{3}}`

(C) `\frac{\sqrt{3}}{2}`

(D) 1

Answer. (A)

Solution.

Since it is given that,

sin `\x` + sin `\y` + sin `\z` = 0

and,                                      cos `\x` + cos `\y` + cos `\z` = 0

Both equations can be written as,

                                             sin `\x` + sin `\y` = -sin `\z`                     .....(1)

and,                                    cos `\x` + cos `\y` = -cos `\z`                    .....(2)

Now we will square both equations (1), (2) and add them

`\sin^{2}x` + `\sin^{2}y` + 2`\sin x` `\sin y` + `\cos^{2}x` + `\cos^{2}y` + 2`\cos x` `\cos y` = `\sin^{2}z` + `\cos^{2}z`

Now, we will use the following trigonometric identity,

`\sin^{2}A` + `\cos^{2}A` = 1

So we will get now,

`\sin^{2}x` + `\cos^{2}x` + `\sin^{2}y` + `\cos^{2}y` + 2( `\sin x` `\sin y`+ `\cos x` `\cos y`) = 1

1 + 1 + 2`\cos`(`\x`-`\y`) = 1

2`\cos`(`\x`-`\y`) = -1

`\cos`(`\x`-`\y`) = `\frac{-1}{2}`

Now we will calculate the value of `\sin`(`\x`-`\y`) using the same identity,

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

`\sin^{2}`(`\x`-`\y`) + `\frac{1}{4}` = 1

`\sin^{2}`(`\x`-`\y`) = `\frac{3}{4}`

`\sin`(`\x`-`\y`) = `\frac{\sqrt{3}}{2}`

Now, `\tan`(`\x`-`\y`) = `\frac{\sin(x-y)}{\cos(x-y)}` = -`\sqrt{3}`

Hence, `\tan`(`\x`-`\y`) = -`\sqrt{3}`

Question 6.) If,

`tan^{-1}``\frac{1}{1-2x}` - `tan^{-1}``\frac{1}{x-1}` = `tan^{-1}``\frac{1}{x^{2}}`

Then, which of the following cannot be the value of x?

(A) -1

(B) 1

(C) 2

(D) 0

Answer. (C)

Solution.

Since it is given that,

`tan^{-1}``\frac{1}{1-2x}` - `tan^{-1}``\frac{1}{x-1}` = `tan^{-1}``\frac{1}{x^{2}}`

Now we will use the following formula,

`tan^{-1}``A` - `tan^{-1}``B` = `tan^{-1}``\frac{A-B}{1+AB}`

`tan^{-1}``\frac{1}{1-2x}` - `tan^{-1}``\frac{1}{x-1}` = `tan^{-1}``\frac{3x-2}{3x-2x^{2}}`

or, `tan^{-1}``\frac{3x-2}{3x-2x^{2}}` = `tan^{-1}``\frac{1}{x^{2}}`

or, `\frac{3x-2}{3x-2x^{2}}` = `\frac{1}{x^{2}}`

or, `3x^{3}` - `2x^{2}` = `3x` - `2x^{2}`

or, `x`(`x^{2}`-1) = 0

So, `x` = -1, 0, 1

Question 7.) If sin `\theta` = cos 2ɸ then find the value of nπ ± ɸ in terms of `\theta`. Where, ‘n' belongs to a set of natural numbers.

(A) `\frac{π}{6}` + `\theta/4`

(B) `\frac{π}{2}` - `\theta`

(C) `\frac{π}{4}` - `\theta/2`

(D) π + 2`\theta`

Answer. (C)

Solution.

Since it is given that,

sin `\theta` = cos 2ɸ

Now here we will use the following relationship,

 sin `\theta` = cos(`\frac{π}{2}` - `\theta`)

So our equation will be,

cos(`\frac{π}{2}` - `\theta`) = cos 2ɸ

cos(`\frac{π}{2}` - `\theta`) = cos(2nπ ± 2ɸ)

`\frac{π}{2}` - `\theta` = 2nπ ± 2ɸ

nπ ± ɸ = `\frac{π}{4}` - `\theta/2`

Question 8.) If `tan` 𝛼 and `tan` 𝛽 are the roots of the following equation,

2`x^{2}` - λ`\sqrt{2}``x` + 1 = 0

Then for what value of λ both the roots are positive. Given that, `tan^{2}`(𝛼+𝛽) = 18 and λ is not equal to 0.

(A) 3

(B) -3

(C) 1

(D) -2

Answer. (A)

Solution.

Since it is given that,

`tan^{2}`(𝛼+𝛽) = 18

Now here we will use the following relationship,

`tan` (A+B) = `\frac{\tanA+\tanB}{1-\tanA\tanB}`

But here we need to calculate the values of `tan` 𝛼 and `tan` 𝛽 first and for that we will use the following equation given in the question,

2`x^{2}` - λ`\sqrt{2}``x` + 1 = 0

Since it is also given that `tan` 𝛼 and `tan` 𝛽 are the roots of the above equations,

Hence sum of the roots will be,

`tan` 𝛼 + `tan` 𝛽 = `\frac{\lambda}{\sqrt{2}}`

And the product of the roots will be,

`tan` 𝛼.`tan` 𝛽 = `\frac{1}{2}`

Now we will evaluate the value of `tan` (𝛼+𝛽) by using above values,

Hence, `tan` (𝛼+𝛽) = `\sqrt{2}`λ

Now we will put this value in this equation, `tan^{2}`(𝛼+𝛽) = 18

Now we will solve it for λ, and we will get λ = ±3

Here we have two values of λ, and we need to choose that value which fulfills the condition that both the roots must be positive. So only positive roots can give a positive sum and a positive product. Hence, λ = 3 satisfies these conditions.

Question 9.) If `\frac{2\sin^{2}A}{1+\cos2A}` = `\frac{1}{49}`

and `\sqrt{\frac{1-\cos2B}{2}}` = `\frac{1}{\sqrt{10}}`.

Then find the value of `tan`(A+B). Given that A, B ∈ [0, `π/2`].

(A) `\frac{3}{10}`

(B) `\frac{1}{2}`

(C) `\frac{4}{3}`

(D) `\frac{2}{9}`

Answer. (B)

Solution.

Since it is given that,

`\frac{2\sin^{2}A}{1+\cos2A}` = `\frac{1}{49}`      .....(1)

and `\sqrt{\frac{1-\cos2B}{2}}` = `\frac{1}{\sqrt{10}}`      .....(2)

Here we will use the following formulas to solve above two equations,

1 + `cos`2A = 2`cos^{2}`A

1 - `cos`2B = 2`sin^{2}`B

Now on solving eq.(1) we will get, `\tan`A = `\frac{1}{7}`

Similarly on solving eq.(2) we will get, `\sin`B = `\frac{1}{\sqrt{10}}`

and, `\cos`B = `\frac{3}{\sqrt{10}}`

Hence, `\tan`B = `\frac{1}{3}`

Now we will use the following formula to evaluate `tan` (A+B)

`tan` (A+B) = `\frac{\tanA+\tanB}{1-\tanA\tanB}`

Now putting values of `\tan`A and `\tan`B and further calculation we will get

`tan` (A+B) = `\frac{1}{2}`

Question 10.) Find the maximum value of `sin` (`theta` + `\phi`) + `cos` (`theta` - `\phi`). Given that `cos``\theta/2`= `\frac{3}{5}`.

(A) `\frac{1}{\sqrt{3}}`

(B) `\frac{23\sqrt{2}}{13}`

(C) `\frac{17\sqrt{3}}{5}`

(D) `\frac{31\sqrt{2}}{25}`

Answer. (D)

Solution.

Since it is given that,

`cos``\theta/2` = `\frac{3}{5}`

Now, we will use the following trigonometric identity  to evaluate `sin``\theta/2`,

`\sin^{2}A` + `\cos^{2}A` = 1

Hence, we will get,

`sin``\theta/2` = `\frac{4}{5}`

Now, we will use the following trigonometric formula to evaluate `sin``\theta` and `cos``\theta`,

`sin``\theta` = 2`sin``\theta/2``cos``\theta/2`

Now putting values of `sin``\theta/2` and `cos``\theta/2` and further calculation we will get,

`sin``\theta` = `\frac{24}{25}` & `cos``\theta` = `\frac{7}{25}`

Now we have to calculate the maximum value of,

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

It can be written as,

`sin``theta``cos``\phi` + `cos``theta``sin``\phi` + `cos``theta``cos``\phi` + `sin``theta``sin``\phi`

Now putting values of `sin``\theta` & `cos``\theta` we will get,

`sin` (`theta` + `\phi`) + `cos` (`theta` - `\phi`) = `\frac{31}{25}`(`sin``\phi`+`cos``\phi`)

Now as we know that the maximum value of x`sin``\phi`+y`cos``\phi` is given by,

`\sqrt{x^{2}+y^{2}}`

Here x = y = 1. Hence, the maximum value will be,

`\frac{31}{25}`(`\sqrt{1^{2}+1^{2}}`) = `\frac{31\sqrt{2}}{25}`

Hence, the maximum value of `sin` (`theta` + `\phi`) + `cos` (`theta` - `\phi`) will be, `\frac{31\sqrt{2}}{25}`