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Calculation:

We know that for range for odd power of sin function always lies between -1 ≤ sinⁿ θ ≤ 1 and

The range for odd power of cosec function always lies between -1 ≥ Cosecn θ ≥ 1

∴ Both statements are correct

Hence, option (3) is correct

Given :

Observer height is 1.6 m 

Height of tower = 20√3

Angle of elevation = 30°

RQ = 20√ 3

Calculation :

In ∆ PQR 

Tanθ = P/B

Tan30° = PQ/RQ

\(\frac{1}{{√ 3 }} = \frac{{PQ}}{{RQ}}\)

PQ = 20 m

Hence total height of tower = 20 + 1.6 = 21.6 m

Concept:

The area of any triangle can be defined as:

A = \(\rm \sqrt{s(s-a)(s-b)(s-c)}\) 

where a, b and c are the sides of the triangle and \(\rm s=\dfrac{a+b+c}{2}\) 

Calculation:

Given \(\rm s=\dfrac{a+b+c}{2}=6\)

Area of the triangle = \(\rm \sqrt{s(s-a)(s-b)(s-c)}\) = Δ

⇒ s(s - a)(s - b)(s - c) = Δ²

Multiplying both side by s 

⇒ s × s(s - a)(s - b)(s - c) = s × Δ2

⇒ s²(s - a)(s - b)(s - c) = 6 Δ2

Multiplying both side by \(\rm 1\over 3\)

⇒ \(\boldsymbol{\rm 1\over 3}\) s2(s - a)(s - b)(s - c) = 2 Δ2

Concept:

\(\rm \tan (α + β) = \dfrac {tan \alpha + tan \beta }{1 -tan \alpha \;tan \beta }\)

Calculations:

Given, α and β are positive angles such that \(α + β = \dfrac{\pi}{4}\), 

⇒\(\rm \tan (α + β) = \tan (\dfrac{\pi}{4})\)

⇒\(\rm \dfrac {tan \alpha + tan \beta }{1 -tan \alpha \;tan \beta } = 1\)

⇒\(\rm {tan \alpha + tan \beta }= 1 -tan \alpha \;tan \beta \)

⇒\(\rm {tan \alpha +tan \alpha \;tan \beta + tan \beta } - 1 = 0 \)

⇒\(\rm {tan \alpha +tan \alpha \;tan \beta + tan \beta } +1-2= 0 \)

⇒\(\rm {tan \alpha +tan \alpha \;tan \beta + tan \beta } +1=2\)

⇒\(\rm tan \alpha (1 +tan \beta ) + (1+tan \beta)=2\)

⇒\(\rm (1+ tan \alpha )(1 +tan \beta )=2\)

Hence, If α and β are positive angles such that \(α + β = \dfrac{\pi}{4}\), then  (1 + tan α) (1 + tan β) equal to 2.  

Concept:

\(\rm sin^2x+cos^2x=1\)

\(\rm \tan^{-1}(tanx)= x\)

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

Calculation:

Let, \(\rm \sin^{-1} \dfrac{4}{5} =x\)

sin x= 4/5

Now, \(\rm cos^2x=1-sin^2x=1-\frac{16}{25}=\frac{9}{25}\)

⇒ cos x = 3/5

Now, cot x = cos x/sin x = 3/4             ....(1)

\(\rm 2 \tan^{-1} \dfrac{1}{3} =tan^{-1}(\frac{\frac 2 3}{1-\frac 19})\)                    ....(∵ \(\rm 2 \tan^{-1} x= tan^{-1}(\frac{2x}{1-x^2})\))

\(\rm =tan^{-1}(\frac{\frac 2 3}{\frac 89})\)

\(\rm =tan^{-1}(\frac 3 4)\)

\(\rm =tan^{-1}(cot x)\)                                  ....(from 1)

\(\rm =tan^{-1}(tan (\frac π 2-x))\)

= π/2 - x                                           .....(∵ \(\rm \tan^{-1}(tanx)= x\))

= \(\rm \frac \pi 2-\sin^{-1} \dfrac{4}{5} \)

∴ \(\sin^{-1} \dfrac{4}{5} + 2 \tan^{-1} \dfrac{1}{3} = \sin^{-1} \dfrac{4}{5}+\frac\pi 2 -\sin^{-1} \dfrac{4}{5}\)

= \(\dfrac{\pi}{2}\)

Concept:

Trigonometric Formulas:

\(\ \rm cosec \;\theta = \dfrac 1{\sin \;\theta}\)

 \(\sin \;(-\theta )= -\sin\;\theta\)

sin (A - B) = sin A. sin B - cos A cos B

Calculations:

Consider,  \(\left(\dfrac{\sec 18^\circ}{\sec 144^\circ} + \dfrac{\text{cosec} \ 18^\circ}{\text{cosec} \ 144^\circ}\right)\) 

=\(\left(\dfrac{ \dfrac 1 {\cos 18^\circ}}{\dfrac 1 {\cos 144^\circ}} + \dfrac{\dfrac 1 {\sin\ 18^\circ}}{\dfrac 1 {\sin \ 144^\circ}}\right)\)

=\(\left(\dfrac{\cos 144^\circ}{\cos 18^\circ} + \dfrac{\text{sin} \ 144^\circ}{\text{sin} \ 18^\circ}\right)\)

=\(\left(\dfrac{\cos (180^\circ - 36^\circ)}{\cos 18^\circ} + \dfrac{\text{sin}( \ 180^\circ-36^\circ)}{\text{sin} \ 18^\circ}\right)\)

=\(\left(\dfrac{\cos 36^\circ}{\cos 18^\circ} - \dfrac{\text{sin}36^\circ}{\text{sin} \ 18^\circ}\right)\)

=\(\left(\dfrac{\cos 36^\circ \sin 18^\circ -{\sin}36^\circ\cos 18^\circ }{\text{sin} \ 18^\circ.\cos 18^\circ}\right)\)

=\(\left(\dfrac{\sin (18^\circ -36^\circ) }{\text{sin} \ 18^\circ.\cos 18^\circ}\right)\)

=\(-\left(\dfrac{\sin 18^\circ }{\text{sin} \ 18^\circ.\cos 18^\circ}\right)\)

=\(-\left(\dfrac{1 }{\cos 18^\circ}\right)\)

= -sec 18° 

Hence,  \(\left(\dfrac{\sec 18^\circ}{\sec 144^\circ} + \dfrac{\text{cosec} \ 18^\circ}{\text{cosec} \ 144^\circ}\right) =-\sec 18^\circ\)

Concept:

In the first quadrant, the values for sin, cos and tan are positive.

In the second quadrant, the values for sin and cosec are positive only.

In the third quadrant, the values for tan are positive only.

In the fourth quadrant, the values for cos are positive only.

sin(π - θ) = sin θ

Calculation:

Here, \(\dfrac{\sin \theta + \cos \theta - \tan \theta}{\sec \theta + \text{cosec} \ \theta - \cot \theta}\)and \(\theta = \dfrac{3\pi}{4} =\pi -\frac \pi 4\)

\(\rm \dfrac{\sin \theta + \cos \theta - \tan \theta}{\sec \theta + \text{cosec} \ \theta - \cot \theta}=\)\(\dfrac{\sin (\pi -\frac \pi 4)+ \cos(\pi -\frac \pi 4) - \tan(\pi -\frac \pi 4)}{\sec (\pi -\frac \pi 4) + \text{cosec} \ (\pi -\frac \pi 4) - \cot (\pi -\frac \pi 4)}\)

\(=\dfrac{\sin (\frac \pi 4)- \cos( \frac \pi 4) + \tan(\frac \pi 4)}{-\sec (\frac \pi 4) + \text{cosec} \ (\frac \pi 4) +\cot (\frac \pi 4)}\)

\(=\frac{\frac{1}{\sqrt2}-\frac{1}{\sqrt2}+1}{-\sqrt2+\sqrt2+1}\)

= 1

Concept:

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

\(\rm sin^2x+cos^2x=1\)

Calculation:

To find: cos{cos-1(3/5) + cos-1(12/13)} = ?

Let, cos-1(3/5) = A ⇒cos A = 3/5

⇒ sin A = \(\sqrt{({1-\frac{3^2}{5^2}})}\)

 = √ (16/25)

= 4/5

Also, let cos-1(12/13) = B ⇒cos B = 12/13

⇒ sin B = \(\sqrt{(1-\frac{12^2}{13^2})}=\sqrt{(\frac{169-144}{169})}\)

 = √(25/169) 

= 5/13

A + B = cos-1(3/5) + cos-1(12/13)

cos{cos-1(3/5) + cos-1(12/13)} = cos(A + B)

Now, we know, cos(A + B) = cos A cos B - sin A sin B 

 \(=\frac{3}{5}\times \frac{12}{13}-\frac{4}{5}\times \frac{5}{13}\\ =\frac{36}{65}-\frac{20}{65}\)

= 16/65

Hence, option (3) is correct.

Identity:

cosec²θ - cot²θ = 1

Calculation:

cosec2θ - cot2θ = 1

⇒ cot²θ - cosec²θ = -1

⇒ 7(-1)

∴ -7

The Correct option is 3 i.e. -7

Given - 

cosθ = (5/13)

Formula used - 

sec²θ  = 1 + tan²θ 

cosec²θ = 1 + cot²θ 

Solution - 

(sec2θ - tan2θ) × (cosec2θ - cot2θ)

⇒ By using above formula,

⇒ 1 × 1 

⇒ 1

Ans = 1.

Given:

Two identities are given

Formula Used:

Basic concept of trigonometric ratio and identities

We know that

(a² – b²) = (a + b)(a - b)

sin²θ + cos²θ = 1

1 + tan²θ = sec²θ

sinθ = 1/cosecθ

Calculation:

The given identity is 1/(1 + cosθ) + 1/(1 - cosθ) = 2cosec²θ

L.H.S.

Take LCM of denominators

⇒ 2/(1 + cosθ) × (1 - cosθ)

⇒ 2/(1 - cos²θ)

⇒ 2/sin²θ

⇒ 2 cosec²θ = R.H.S.

So, statement 1 is correct

Now, the given identity is (1 + tan²θ)/ cosec²θ = tan²θ

L.H.S.

⇒ (1 + tan²θ)/cosec²θ

⇒ sec²θ/cosec²θ

⇒ sin²θ/cos²θ

⇒ tan²θ = R.H.S.

So, statement 2 is correct

Hence, option (3) is correct

Given:

sinθ + cosecθ = 2

Formula used:

sin90° = 1

cos90° = 0

cosec90° = 1

sinθ = 1/cosecθ 

Calculation:

sinθ + cosecθ = 2

⇒ sinθ + (1/sinθ) = 2

But the above equation possible when sinθ = 1

⇒ sinθ = sin90° 

After comparing we get θ = 90° 

sin2θ + cos4θ - 2cosec2θ

⇒ sin290° + cos490° - 2cosec290°

⇒ (1)² + (0)⁴ - 2 × (1)²

⇒ 1 + 0 - 2

⇒ -1

∴ The value is -1.

Given:

tanθ + cotθ = √3

Formula used:

a³ + b³ = (a + b)³ - 3ab(a + b)

a² + b² = (a + b)² - 2(a × b)

tanθ × cotθ = 1

Calculation:

tanθ + cotθ = √3

Taking cube on both sides, we get

(tanθ + cotθ)³ = (√3)³

⇒ tan³θ + cot³θ + 3 × tanθ × cotθ × (tanθ + cotθ) = 3√3

⇒ tan3θ + cot3θ + 3√3  = 3√3

⇒ tan3θ + cot3θ = 0  

Taking square on the both sides

(tan3θ + cot3θ)² = 0

⇒ tan⁶θ + cot⁶θ + 2 × tan3θ × cot3θ = 0

⇒ tan6θ + cot6θ + 2 = 0    

⇒ tan6θ + cot6θ = - 2

∴ The value of tan6θ + cot6θ is - 2.

Given:

tanθ = 5/12

Calculation:

(cosθ - sinθ)/(cosθ + sinθ)

After taking cosθ  common from both numerator and denominator we get,

⇒ [1 - (sinθ/cosθ)]/[1 + (sinθ/cosθ)] 

⇒ [1 - tanθ]/[1 + tanθ] 

⇒ [1 - (5/12)]/[1 + (5/12)] 

⇒ [(12 - 5)/12]/[(12 + 5)/12]

⇒ [7/12]/[17/12]

⇒ 7/17

∴ The value is 7/17.

Given:

Our equation is (secθ – tanθ) = 2

Formula used:

sec²θ – tan²θ = 1

Concept used:

(secθ – tanθ)(secθ + tanθ) = 1

If (secθ – tanθ) = p, then (secθ + tanθ) = 1/p

Calculation:

Our given expression is (secθ – tanθ) = 2      ----(1)

⇒ (secθ + tanθ) = ½      ----(2)

Adding equation (1) and (2)

⇒ 2secθ = 2 + 1/2

⇒ secθ = 5/4

∴ The value of cosθ is 4/5

Given:

cosθ = 3/5

Concept used:

In a right-angled triangle

cosθ = base/hypotenuse

tanθ = perpendicular/base

Pythagorus theorem

(Hypotenuse)² = (Perpendicular)² + (Base)²

Calculation:

Base = 3, Hypotenuse = 5

By using pythagorus theorem

⇒ (Perpendicular)² = (5)² – (3)²

⇒ (Perpendicular)² = 25 – 9

⇒ (Perpendicular)² = 16

⇒ Perpendicular = 4

tanθ = perpendicular/base

⇒ tanθ = 4/3

Given 

√3cotθ = 1 

Calculation 

cotθ = 1/√3 

⇒ θ = 60° 

⇒ cos²θ - sin²θ 

⇒ cos²60 - sin²60 

⇒ (1/2)² - (√3/2)²

⇒ 1/4 - 3/4 = -2/4

⇒ -1/2 

∴ The value of cos²θ - sin²θ is -1/2

Given:

sinθ = 0.96

Formula used:  

sinθ = perpendicular/hypotenuse,

cotθ = base/perpendicular

By Pythagoras theorem,

Hypotenuse2 = Perpendicular2 + Base2

Calculation:

According to the question:

sinθ = 0.96 = 96/100 = 24/25

Perpendicular = 24

⇒ Hypotenuse = 25

Hypotenuse2 = Perpendicular2 + Base2

⇒ 252 = 242 + Base2

⇒ 625 = 576 + Base2

⇒ 625 – 576  = Base2

⇒ Base2 = 49

⇒ Base = √49

⇒ Base  = √(7 × 7) = 7

Now, cotθ = base/perpendicular

⇒ 7/24 = 0.2916

∴ The value of cotθ = 0.2916.

Given

sin 3θ = cos(θ - 26°)

Formula Used 

sinθ = cos(90° - θ)

cos θ = sin(90° - θ)

Calculation 

sin 3θ = sin [90° - (θ - 26°)]

⇒ 3θ = 90° - (θ - 26°)

⇒ 4θ = 116° 

⇒ θ = 29° 

∴ The answer is 29° 

Formula Used:

sin(A – B) = sinA × cosB – sinB × cosA

Calculation:

Now, from the above used formula

⇒ sin15° = sin(45° – 30°)

⇒ sin45° × cos30° – sin30° × cos45°

⇒ (1/√2) × (√3/2) – (1/√2) × (1/2)

⇒ (√3 – 1)/(2 × √2)

∴  (√3 – 1 )/ 2√2

Given:

Cos x = 3/5

Formula used: 

Sin x = perpendicular/hypotenuse,

Cos x = base/hypotenuse,

By Pythagoras theorem,

Hypotenuse2 = Perpendicular2 + Base2

Calculation:

According to the question:

Cos x = 3/5 

Base = 3

⇒ Hypotenuse = 5

Hypotenuse2 = Perpendicular2 + Base2

⇒ 52 = 32 + Perpendicular2

⇒ 25 = 9 + Perpendicular2

⇒ 25 – 9 = Perpendicular2

⇒ Perpendicular2 = 16

⇒ Perpendicular = √16 

⇒ Perpendicular2 = √(4 × 4)

⇒ Perpendicular = 4

Now,

(Sin x – Sin³ x)

⇒ [(perpendicular/hypotenuse) – (perpendicular/hypotenuse)³]

⇒ [4/5 – (4/5)³]

⇒ [4/5 – 64/125]

⇒ (100 – 64)/125

⇒ 36/125 = 0.288

∴ The answer is 0.288.

Concept used:

cosec(90° – x)  = sec x

tan(90° – x) = cot x

Calculation:

⇒ cosec(75° + x) – sec(15° – x) + tan(55° + x) – cot(35° – x)

⇒ cosec[90° – (15° - x)]  – sec(15° – x) + tan[90° – (35° – x)] – cot(35° – x)

⇒ sec(15° – x)  – sec(15° – x) + cot(35° – x) – cot(35° – x)

⇒ 0 + 0 = 0

∴ The value of cosec(75° + x) – sec(15° – x) + tan(55° + x) – cot(35° – x) is 0

(a) Here, sin(A + 2B) = √3/2 

sin(A + 2B) = sin 60° 

A + 2B = 60° ...(i) 

and cos(A + 4B) = 0 

cos (A + 4B) = cos 90° 

A + 4B = 90° 

A = – 4B + 90° ...(ii) 

Put the value of A in eq. (ii), 

A + 2B = 60° 

or – 4B + 2B + 90° = 60° 

– 2B = 60° – 90° 

– 2B = – 30° 

B = 15° 

and A = – 4B + 90° 

= – 4(15°) + 90° 

A = 30°

Given:

It is given that a cos 60° = b cosec 45°

Formula Used:

Basic concept of trigonometric ratio and identities

We know that

Cos 60° = 1/2 and cosec 45° = √2

Calculation:

We have to find the value of 7(a/2b)² = 7a²/4b²

By rearranging the terms we get

∴ a/b = cosec 45°/ Cos 60°

⇒ a/b = 2√2

Now, squaring both the sides

∴ (a/b)² = (2√2)²

⇒ a²/b² = 8

Now, multiply both the side by 7/4

∴ 7/4 × a²/b² = 7/4 × 8 = 14

Hence, option (1) is correct

Concept:

Trigonometry: \(\rm \cos 2\theta =\frac{1-{{\tan }^{2}}\theta }{1+{{\tan }^{2}}\theta }\).

Calculation:

Since a, b and c are in A.P., we have 2b = a + c ⇒ c = 2b - a.

It is given that \(\rm \cos \alpha =\frac{a}{b+c}\).

Using the half-angle formula \(\rm \cos 2\theta =\frac{1-{{\tan }^{2}}\theta }{1+{{\tan }^{2}}\theta }\) and the fact that c = 2b - a, we get:

⇒ \(\rm \frac{1-{{\tan }^{2}}\frac{\alpha }{2}}{1+{{\tan }^{2}}\frac{\alpha }{2}}=\frac{a}{3b-a}\)

⇒ \(\rm \frac{2{{\tan }^{2}}\tfrac{\alpha }{2}}{2}=\frac{(3b-a)-a}{(3b-a)+a}\)

⇒ \(\rm {{\tan }^{2}}\frac{\alpha }{2}=\frac{3b-2a}{3b}\)            ...(1)

Similarly, \(\rm {{\tan }^{2}}\frac{\gamma }{2}=\frac{2a-b}{3b}\)            ...(2)

Now, \(\rm {{\tan }^{2}}\frac{\alpha }{2}+{{\tan }^{2}}\frac{\gamma }{2}\)

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

= \(\rm \frac{2b}{3b}\)

= \(\frac{2}{3}\)

Concept:

\(\rm cosA + cosB = 2cos(\frac{A+B}{2})cos(\frac{A-B}{2})\)

\(\rm cosA - cosB = -2sin(\frac{A+B}{2})sin(\frac{A-B}{2})\)

Calculation:

Here, \(\rm \frac{\cos(x+y)}{\cos(x-y)}=\frac{a+b}{a-b}\)

Applying componendo and dividendo, we get 

\(\rm \frac{cos(x+y)+cos(x-y)}{cos(x+y)-cos(x-y)}=\frac{a+b + a-b}{a+b-a + b}\)

\(⇒ \rm \frac{2cos(\frac{x+y+x-y}{2})cos(\frac{x+y-x+y}{2})}{-2sin(\frac{x+y+x-y}{2})sin(\frac{x+y-x+y}{2})}=\frac{2a}{2b}\)

⇒ cot(x) cot(y) = -a/b

⇒ tan(x) tan(y) = -b/a

Hence, option (3) is correct.

Identity used:

Sin²A = 1 – Cos²A 

Calculation:

x = {[Sec A/(1 – Cos A)] × (1 + CosA)/(1 + CosA)

⇒  x = (Sec A + Sec A Cos A)/(1 – CosA) × (1 + CosA)

⇒ x = (SecA + 1)/(1 – Cos²A)

⇒  x = (Sec A + 1)/Sin²A

⇒  1/x = Sin²A/(Sec A + 1)

Given:

tan24.tan48.tan42.tan66

Formula used:

tanθ = cot(90 – θ)

tanθ = 1/(cotθ)

Calculation:

tan48 = cot(90 – 48) = cot42

tan66 = cot (90 – 66) = cot24

Now we have

tan24.cot24.tan42.cot42

⇒ tan24.(1/tan24).tan42.(1/tan42) = 1

∴ The value of tan24.tan48.tan42.tan66 is 1.