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Given:

SinA = 1/x

Concept Used:

Sinθ = Perpendicular / Hypotenuse

Cosθ = Base / Hypotenuse

Base2 + Perpendicular2 = Hypotenuse²

Calculation:

Sinθ = Perpendicular / Hypotenuse

SinA = 1/x

Comparing get,

Perpendicular = 1 and Hypotenuse = x

Base = √x² - 1

CosA = Base / Hypotenuse

⇒ CosA = \({\sqrt{x^2-1} \over x}\)

∴ CosA = \({\sqrt{x^2-1} \over x}\).

Given:

tanα = a/b

Concept Used:

tanθ = Perpendicular/Base

Sinθ = Perpendicular/Hypotenuse

Cosθ = Base/Hypotenuse

Base² + Perpendicular² = Hypotenuse²

Calculation:

tanα = a/b

and tanα = Perpendicular/Base

Comparing above two get,

Perpendicular = a and Base = b

Hypotenuse = √a² + b²

Sinα = Perpendicular/Hypotenuse

⇒ Sinθ = a/√a² + b²

Cosα = Base/Hypotenuse

⇒ Cosα = b/√a2 + b2

Sinα + Cosα =  (a/√a2 + b²) + (b/√a2 + b²)

⇒ Sinα + Cosα = \({a + b} \over \sqrt{a^2 + b^2}\)

∴ Sinα + Cosα = \({a + b} \over \sqrt{a^2 + b^2}\)

Calculation:

Given cot θ + tan θ = 2

\(\rm {1\over \tanθ} +\tan θ = 2\)

tan² θ - 2 tan θ + 1 = 0

(tan θ - 1)² = 0

tan θ = 1

θ = \({\pi\over4}, {5\pi\over4}, {9\pi\over4}...\) = \(\frac{{n\pi }}{2} + {( - 1)^n}\frac{\pi }{4}\), (n ∈ N)

Solution: 

Explanation: Use one of the Pythagorean identity namely,

sec²A=1+tan²A

LHS = sec⁴A−sec²A
=(sec²A)²−sec²A
=(1+tan²A)²−(1+tan²A)
=1+2tan²A+tan⁴A−1−tan²A

=tan⁴A+tan²A

= RHS

Concept:

sin (A + B) = sin A cos B + cos A sin B

cos2 x =  1 - sin2 x

cos 2x = 1 - 2sin2 x

Calculations:

Consider, sin 3x 

⇒sin (2x + x)

⇒ sin 2x cos x + cos 2x sin x

⇒ 2 sin x cos x. cos x + (1 - 2sin² x) sin x

⇒ 2 sin x cos² x + (1 - 2sin2 x) sin x

⇒ 2 sin x (1 - sin2 x) + (sin x - 2sin3 x)

⇒ 2 sin x  - 2sin3 x + sin x - 2sin3 x

⇒ 3 sin x  - 4sin3 x 

Hence, sin 3x = 3 sin x  - 4sin3 x 

Concept:

Trigonometry Formula:

sin (-θ) = - sin θ

sin (360 + θ) = sin θ

sin 90° = 1

Calculation:

sin (-450°)

= - sin 450°        [∵ sin (-θ) = - sin θ]

= - sin (360° + 90°)

= - sin 90°          [∵ sin (360 + θ) = sin θ]

= -1

Concept:

2sinx cos x = sin 2x

sin (90° + x) = cos x

Calculation:

Given, 2 sin 75° cos 75°

= sin [2 (75°)]

= sin (150°)

= sin (90° + 60°)

= cos 60°

= \(\rm \dfrac 1 2\)

Hence, the value of 2 sin 75° cos 75° is \(\rm \dfrac 1 2\)

Given:

Angle 45° 45' is a complementry angle

Concept used:

Sum of complementary angles is 90° 

1° = 60 minutes

Calculation:

Let the other angle is x

⇒ x + 45° 45' = 90° 

⇒ x = 90° – 45° 45'

⇒ x = 44° 15'

∴ Complementry angle of 45° 45' is 44° 15'.

Concept:

2cos2 θ  - 1 = cos 2θ 

Calculations:

Consider, 2cos2 15° - 1

= cos 2 (15°) 

= cos (30°) 

= \(\dfrac {\sqrt 3}2\)

Given:

Two statements are given

Formula Used:

Basic concept of trigonometric ratio and identities

We know that if A, B and C is the angles of triangle or their sum is 180°

∴ tan (A + B)/2 = cot (C/2)

∴ sin (A + B)/2 = cos (C/2)

∴ sec (A + B)/2 = cosec (C/2)

Where versa vice is possible

Calculation:

P, Q and R are the angles of a triangle

∴ tan (P + Q)/2 = cot (R/2)

But it is given in statement I

tan (P + Q)/2 = sec (R/2)

∴ Statement 1 is incorrect

Now, X + Y + Z = 180°

∴ Sin (X + Y)/2 = Cos (Z/2)

So, Statement 2 is correct

Hence, option (2) is correct

Given:

It is given that sec (θ + 45) ° = √8/2

Formula Used:

Basic concept of trigonometric ratio and identities

We know that

Sec 45° = √2 and tan 0° = 0

Calculation:

We can simplifies the given expression

∴ Sec (θ + 45) ° = 2√2/2

⇒ Sec (θ + 45) ° = √2

⇒ Sec (θ + 45) ° = Sec 45°

⇒ (θ + 45) ° = 45°

⇒ θ = 0°

Now, we have to find tan θ

∴ tan0° = 0

Hence, option (1) is correct

Concept

Sin²A + Cos²A = 1

Calculation

Statement 1

Cos²(90°– 84°) + Cos² (90° – 70°) + Cos²84° + Cos² 70°

⇒  Sin²84° + Sin²70° + Cos²84 °+ Cos²70°

⇒  Sin²84 °+ Cos²84 °+ Sin²70°+ Cos²70°

⇒  1 + 1

⇒  2

Statement 2

Sin 45° = (1/√2)

Cos 45° = (1/√2)

Statement 3

Sin 0° + Cos 0°

⇒   0 + 1

⇒   1

Hence, every statement is right

Given:

tan(210°)

Concept used:

tan(180° + θ) = tanθ, tan is positive in 3rd quadrant.

Calculation:

tan(210°) = tan(180° + 30°)

⇒ tan(210°) = tan30° 

⇒ tan(210°) = 1/√3

∴ The value of tan(210°) is 1/√3.

Given:

Cos 225°

Formula Used:

Cos (180° + θ ) = -Cos θ

Concept Used:

Cosθ  in the third quadrant gives negative value.

Cos 45° = 1/√2

√2 = 1.414

Calculation: 

Cos 225° = Cos (180° + 45° )

⇒ Cos 225° = -Cos 45°

⇒ -Cos 45° = -1/√2 = 1000/1414 = -0.7071

∴ The value of Cos 225° is -1/√2 or -0.7071

Concept:

It is known that \({\cot ^{ - 1}}\left( { - x} \right) = \pi - {\cot ^{ - 1}}\left( x \right)\) for all value \(x \in R\).

Calculation:

The expression be rewritten as \({\cot ^{ - 1}}\left( {\frac{4}{5}} \right) + {\cot ^{ - 1}}\left( { - \frac{4}{5}} \right)\)

\(= {\cot ^{ - 1}}\left( {4/5} \right) + \left( {\pi - {{\cot }^{ - 1}}\left( {4/5} \right)} \right)\)

\(= {\cot ^{ - 1}}\left( {4/5} \right) + \pi - {\cot ^{ - 1}}\left( {4/5} \right)\)

\(= \pi\)

Concept:

\({\cos ^{ - 1}}\left( {\sqrt {1 - {x^2}} } \right) = {\tan ^{ - 1}}\left( {\frac{x}{{\sqrt {1 - {x^2}} }}} \right)\)

\({\tan ^{ - 1}}x - {\tan ^{ - 1}}y = {\tan ^{ - 1}}\left( {\frac{{x - y}}{{1 + xy}}} \right)\)

Calculation:

As we know, \({\cos ^{ - 1}}\left( {\sqrt {1 - {x^2}} } \right) = {\tan ^{ - 1}}\left( {\frac{x}{{\sqrt {1 - {x^2}} }}} \right)\)

⇒ \({\cos ^{ - 1}}\frac{4}{5} = {\tan ^{ - 1}}\frac{3}{4}\;\ and\ {\cos ^{ - 1}}\frac{{15}}{{17}} = {\tan ^{ - 1}}\frac{8}{{15}}\)

The expression \({\cos ^{ - 1}}\frac{4}{5} - {\cos ^{ - 1}}\frac{{15}}{{17}}\)  can be rewritten as follows:

\({\cos ^{ - 1}}\frac{4}{5} - {\cos ^{ - 1}}\frac{{15}}{{17}} = {\tan ^{ - 1}}\frac{3}{4} - {\tan ^{ - 1}}\frac{8}{{15}}\)

As we know that, \({\tan ^{ - 1}}x - {\tan ^{ - 1}}y = {\tan ^{ - 1}}\left( {\frac{{x - y}}{{1 + xy}}} \right)\)

\({\tan ^{ - 1}}\frac{3}{4} - {\tan ^{ - 1}}\frac{8}{{15}} = {\tan ^{ - 1}}\left( {\frac{{\frac{3}{4} - \frac{8}{{15}}}}{{1 + \frac{2}{5}}}} \right)\)

\(= {\tan ^{ - 1}}\left( {\frac{{13}}{{84}}} \right)\)

Concept:

• \({\cot ^{ - 1}}\left( {\cot x} \right) = x\)
• \(\cot \left( {x + y} \right) = \frac{{\cot x\cot y - 1}}{{\cot y + \cot x}}\)

Calculation:

The expression \({\cot ^{ - 1}}\left( {\frac{{1 - \sin x}}{{\cos x}}} \right)\) can be written as follows:

\({\cot ^{ - 1}}\left( {\frac{{1 - \sin x}}{{\cos x}}} \right) = {\cot ^{ - 1}}\left( {\frac{{{{\cos }^2}\left( {x/2} \right) + {{\sin }^2}\left( {x/2} \right) - 2\sin \left( {x/2} \right)\cos \left( {x/2} \right)}}{{{{\cos }^2}\left( {x/2} \right) - {{\sin }^2}\left( {x/2} \right)}}} \right)\)

\( = {\cot ^{ - 1}}\left( {\frac{{{{\left( {\cos \left( {x/2} \right) - \sin \left( {x/2} \right)} \right)}^2}}}{{\left( {\cos \left( {x/2} \right) - \sin \left( {x/2} \right)} \right)\left( {\cos \left( {x/2} \right) + \sin \left( {x/2} \right)} \right)}}} \right)\)

\(= {\cot ^{ - 1}}\left( {\frac{{\left( {\cos \left( {x/2} \right) - \sin \left( {x/2} \right)} \right)}}{{\left( {\cos \left( {x/2} \right) + \sin \left( {x/2} \right)} \right)}}} \right)\)

Dividing the numerator and denominator of RHS by sin(x/2) , we get

\({\cot ^{ - 1}}\left( {\frac{{1 - \sin x}}{{\cos x}}} \right) = {\cot ^{ - 1}}\left( {\frac{{\left( {cot(\frac{x}{2})\ cot \frac{\pi}{4} - \ 1} \right)}}{{\left( {cot(\frac{\pi}{4} + \cot \left( {\frac{x}{2}} \right)} \right)}}} \right)\)

As we know that, \(\cot \left( {x + y} \right) = \frac{{\cot x\cot y - 1}}{{\cot y + \cot x}}\)

\(= {\cot ^{ - 1}}\left( {\cot \left( {\frac{\pi }{4} + \frac{x}{2}} \right)} \right)\)

\(= \frac{\pi }{4} + \frac{x}{2}\)

Concept:

\({\cos ^{ - 1}}\left( {\cos x} \right) = x\)

Calculation:

\(\cos \left( {{{\sin }^{ - 1}}\frac{1}{5}} \right)\)

Let,

\({\sin ^{ - 1}}\frac{1}{5} = x\)

\(\sin x = \frac{1}{5}\)

We know,

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

\({\left( {\frac{1}{5}} \right)^2} + {\cos ^2}x = 1\)

\({\cos ^2}x = 1 - {\left( {\frac{1}{5}} \right)^2}\)

\({\cos ^2}x = \frac{{24}}{{25}}\)

\(\cos x = \frac{{2\sqrt 6 }}{5}\)

\(x = {\cos ^{ - 1}}\frac{{2\sqrt 6 }}{5}\)

Now substitute  for \({\sin ^{ - 1}}\frac{1}{5}\) in \(\cos \left( {{{\sin }^{ - 1}}\frac{1}{5}} \right)\)

\(\cos \left( {{{\sin }^{ - 1}}\frac{1}{5}} \right) = \cos \left( {{{\cos }^{ - 1}}\frac{{2\sqrt 6 }}{5}} \right)\)

\( = \frac{{2\sqrt 6 }}{5}\)

Concept:

\({\sin ^{ - 1}}x + {\cos ^{ - 1}}x = \frac{\pi }{2}\) and \({\sin ^{ - 1}}\left( { - x} \right) = - {\sin ^{ - 1}}x\)

Calculation:

\({\sin ^{ - 1}}\left( {1 - x} \right) + 2{\sin ^{ - 1}}\left( {2x} \right) + {\cos ^{ - 1}}\left( {2x} \right) = \frac{\pi }{2}\)

\({\sin ^{ - 1}}\left( {1 - x} \right) + 2{\sin ^{ - 1}}\left( {2x} \right) + {\cos ^{ - 1}}\left( {2x} \right) = {\sin ^{^{ - 1}}}\left( {2x} \right) + {\cos ^{ - 1}}\left( {2x} \right)\)

\({\sin ^{ - 1}}\left( {1 - x} \right) + {\sin ^{ - 1}}2x = 0\)

\({\sin ^{ - 1}}\left( {1 - x} \right) = - {\sin ^{ - 1}}\left( {2x} \right)\)

\({\sin ^{ - 1}}\left( {1 - x} \right) = {\sin ^{ - 1}}\left( { - 2x} \right)\)

\(1 - x = - 2x\)

\(x = - 1\)

But as we can see that, by substituting x = - 1 in the given equation we get, sin^-1(2) + sin^-1(-2) = 0, but we know that domain of sin^-1 x is [-1, 1]

So, x = - 1 is not possible solution of the given equation.

Hence, there is no value of x which satisfies the given equation

Concept:

If x = sin θ then \(θ = {\sin ^{ - 1}}x\)

Similarly, if x = tan θ then \(θ = {\tan ^{ - 1}}x\)

Calculation: 

The expression \({\tan ^{ - 1}}\left[ {2\cos \left( {2{{\sin }^{ - 1}}\frac{{\sqrt 3 }}{2}} \right)} \right]\) can be rewritten as:

\({\tan ^{ - 1}}\left[ {2\cos \left( {2{{\sin }^{ - 1}}\frac{{\sqrt 3 }}{2}} \right)} \right] = {\tan ^{ - 1}}\left[ {2\cos \left( {\frac{{2\pi }}{3}} \right)} \right]\)

\(= {\tan ^{ - 1}}\left[ {2\left( { - \frac{1}{2}} \right)} \right]\)

\(= {\tan ^{ - 1}}\left( { - 1} \right)\)

\(= - \frac{\pi}{4}\)

\(2{\tan ^{ - 1}}\left[ {2\cos \left( {2{{\sin }^{ - 1}}\frac{{\sqrt 3 }}{2}} \right)} \right]\) \(= - 2 \times \frac{\pi}{4} = - \frac{\pi}{2}\)

Concept:

It is known that, \({\sin ^{ - 1}}x + {\cos ^{ - 1}}x = \frac{\pi }{2},x \in \left[ { - 1,1} \right]\)

Calculation:

\(y = {\cos ^{ - 1}}\left( {4{x^3} - 3x} \right) + 3{\sin ^{ - 1}}x\)---------(1)

Take \({\cos ^{ - 1}}\left( {4{x^3} - 3x} \right)\)

Let us substitute x = cos θ in \({\cos ^{ - 1}}\left( {4{x^3} - 3x} \right)\)

\({\cos ^{ - 1}}\left( {4{x^3} - 3x} \right) = {\cos ^{ - 1}}\left( {4{{\cos }^3}θ - 3\cos θ } \right)\)

\(= {\cos ^{ - 1}}\left( {\cos 3θ } \right)\)

\(= 3θ \)

\(= 3{\cos ^{ - 1}}x\)

Now substitute the value of \({\cos ^{ - 1}}\left( {4{x^3} - 3x} \right)\) in equation (1)

\(y = {\cos ^{ - 1}}\left( {4{x^3} - 3x} \right) + 3{\sin ^{ - 1}}x\)

\( = 3{\cos ^{ - 1}}x + 3{\sin ^{ - 1}}x\)

As we know that, \({\sin ^{ - 1}}x + {\cos ^{ - 1}}x = \frac{\pi }{2},x \in \left[ { - 1,1} \right]\)

\( y = \frac{3\pi}{2}\)

Concept:

\(\arctan x = y ⇒ x = \tan y\)

Calculation:

Here, we have to find the value of \(\arctan \left( {1/\sqrt 3 } \right) + \operatorname{arcsec} \left( { - 2} \right)\)

\(\tan x = \frac{1}{{\sqrt 3 }}\) ⇒ \(x = \frac{\pi }{6}\)

Also, \(\sec y = -2\) ⇒ \(y = \frac{2\pi }{3}\)

So, \(\arctan \left( {1/\sqrt 3 } \right) + {\mathop{\rm arcsec}\nolimits} \left( { - 2} \right) = x + y\)

\(= \frac{\pi }{6} + \frac{2\pi }{3}\)

\(= \frac{\pi+4\pi}{6} = \frac {5\pi}{6} \)

Sin (A + B) = 1/2
A + B = sin^-1(1/2) 
A + B = 30° ...... (1)

Now,

2 Cos(A + B) = 1
Cos(A + B) = 1/2
A + B = Cos^-1(1/2)
A + B = 60°.........(2)

Equation 1 and 2 contradicts each other.  Hence value of A & B cant be determined from the above two equations.

Concept:

Trigonometric Identities:

\(\rm \tan(A\pm B)=\frac{\tan A \pm \tan B}{1 \mp \tan A \tan B}\).

Calculation:

Given that \(\rm x+y=\frac{\pi}{4}\).

⇒ \(\rm \tan(x+y)=\tan \frac{\pi}{4}\)

⇒ \(\rm \frac{\tan x + \tan y}{1 -\tan x \tan y}=1\)

⇒ tan x + tan y = 1 - tan x tan y            ... (1)

Now, (1 + tan x)(1 + tan y)

= 1 + tan y + tan x + tan x tan y

Using the value in equation (1), we get:

= 1 + 1 - tan x tan y + tan x tan y

= 2.

Concept:

Trigonometric Identities:

• ​\(\rm \cot\theta=\frac{1}{\tan\theta}\).
• \(\rm \cot(A\pm B)=\frac{\mp1+cot A\cot B}{\pm\cot A+\cot B}\).

Calculation:

Given: tan A - tan B = x              ... (1)

And, cot A - cot B = y              ... (2)

⇒ \(\rm \frac{1}{\tan A}-\frac{1}{\tan B}=y\)

⇒ \(\rm \frac{\tan B-\tan A}{\tan A\tan B}=y\)

⇒ \(\rm \frac{-x}{\tan A\tan B}=y\)              ... [Using (1).]

⇒ \(\rm \cot A\cot B=-\frac{y}{x}\)              ... (3)

Now, \(\rm \cot(A- B)=\frac{+1+cot A\cot B}{-\cot A+\cot B}\)

= \(\rm \frac{+1-\frac yx}{-y}\)              ... [Using (2) and (3).]

= \(\rm \frac{y-x}{xy}\)

= \(\rm \frac{1}{x}-\frac{1}{y}\).

Given

2 × cos(A + B) = 2 × sin(A - B) = 1

Concept

Here all three identities are equal to each other so, take first any two identities then the other two identities.

Calculation

Let 2 × cos(A + B) = 1

⇒ cos(A + B) = (1/2)

⇒  cos(A + B) = cos 60° 

⇒ A + B = 60°     ....(1)

Now,

Let 2 × sin(A - B) = 1

⇒ sin(A - B) = (1/2)

⇒ sin(A - B) = sin30° 

⇒ A - B = 30°     ....(2)

Now, add equation (1) and (2)

⇒ (A + B) + (A - B) = 60° + 30° 

⇒ 2A = 90° 

⇒ A = (90°/2)

⇒ A = 45° 

Now put the value of A in equation (1)

⇒ 45° + B = 60° 

⇒ B = 60° - 45° 

⇒ B = 15° 

∴ A = 45° and B = 15° 

Concept:

Trigonometric Identities:

• sin² θ + cos² θ = 1.
• sin 2θ = 2 sin θ cos θ.
• cos 2θ = cos² θ - sin² θ.

Calculation:

Given: sin θ + cos θ = 1

Squaring both sides, we get:

⇒ sin2 θ + cos2 θ + 2 sin θ cos θ = 1

⇒ 1 + sin 2θ = 1

⇒ sin 2θ = 0

Using sin2 θ + cos2 θ = 1, we can say:

sin2 2θ + cos2 2θ = 1

⇒ 0 + cos2 2θ = 1

⇒ cos 2θ = ±1.

Trigonometric Identities:

• sin (A ± B) = sin A cos B ± sin B cos A.
• cos (A ± B) = cos A cos B ∓ sin A sin B.

We know that  sin(x)² + cos(x)² = 1 

(3/5)² + cos(x)² = 1 
9/25 + cos(x)² = 25/25 
cos(x)² = 16/25 
cos(x) = +/- (4/5) 
Since it is in second quadrant cos(x) = -4/5 
sin(y)² + (-12/13)² = 1 
sin(y)² + 144/169 = 169/169 
sin(y)² = 25/169 
sin(y) = +/- 5/13 
Since it is in second quadrant sin(y) = 5/13 
sin(x) = 3/5 
cos(x) = -4/5 
sin(y) = 5/13 
cos(y) = -12/13 
sin(x + y) => sin(x)cos(y) + sin(y)cos(x) 

=> (3/5) * (-12/13) + (5/13) * (-4/5) 

=> -36 / 65 + -20 / 65

=> -56/65

So, sin(x+y) = -56/65

We know that 
sin (x + y) = sin x cos y + cos x sin y ... (1) 
Now cos²x = 1 – sin²x = 1 – 9/25 = 16/25 
Therefore cos x = ± 4/5. 
Since x lies in second quadrant, cos x is negative. 
Hence cos x = −4/5 
Now sin²y = 1 – cos²y = 1 – 144/169 = 25/169 
i.e. sin y = ± 5/13. 
Since y lies in second quadrant, hence sin y is positive. 

Therefore, sin y =5/13. 

Substituting the values of sin x, sin y, cos x and cos y in (1), we get 
sin(x + y) 3/5 × (-12/13) + (−4/5) × 5/13 = (-36/65) –(20/65) = -56/65

Therefore, sin(x+y) = -56/65