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On substituting the values of various T-ratios, we get: 

cos45° cos30° + sin45° sin30°

\((\frac{1}{\sqrt2}\times\frac{\sqrt3}2+\frac{1}{\sqrt2}\times\frac{1}2)\) = \((\frac{\sqrt3}{2{\sqrt{2}}}+\frac{1}{2\sqrt{2}})\) = \((\frac{{\sqrt{3}}+1}{2\sqrt{2}})\)

(b) \(\frac34\)

 \(\sqrt{\frac{(1-sin\,x)(1+sin\,x)}{(1+cos\,x)(1-cos\,x)}}\) = \(\sqrt{\frac{1-sin^2\,x}{1-cos^2\,x}}\)      (∵ (a – b) (a+b) = a² – b²)

= \(\sqrt{\frac{cos^2\,x}{sin^2\,x}}\)                  (∵ sin²x + cos²x = 1)

= \(\frac{cos\,x}{sin\,x}\) = cot θ = \(\frac{1}{tan\,\theta} = \frac{1}{\frac{4}{3}}=\frac34.\)

On substituting the values of various T-ratios, we get: 

cos60° cos30°− sin60° sin30°

= \((\frac{1}2\times\frac{\sqrt{3}}2-\frac{\sqrt{3}}2\times\frac{1}2)\) = \((\frac{\sqrt{3}}4-\frac{\sqrt{3}}4)\) = 0

sin 45° + cos 45°

= \(\frac{1}{\sqrt{2}}\) + \(\frac{1}{\sqrt{2}}\)

= \(\frac{1+1}{\sqrt{2}}\)

= \(\frac{2}{\sqrt{2}}\)

= \(\frac{\sqrt{2}\times\sqrt{2}}{\sqrt{2}}\)

= \(\sqrt{2}\)

We have, 

Tan A = 1 

⇒ sinA/cosA = 1 

⇒ sin A = cos A

⇒ sin A − cos A = 0 

Squaring both sides, we get 

(sinA − cosA)² = 0 

⇒ sin²A + cos²A − 2 sin A . cos A = 0 

⇒ 1 − 2 sin A . cos A = 0 

∴ 2 sin A . cosA = 1

a cos θ – b sin θ = c 

⇒ (a cos θ – b sin θ)² = c² 

(i.e) a² cos² θ + b² sin² θ – 2ab sin θ cos θ = c² 

(i.e) a² (1 – sin² θ) + b² (1 – cos² θ) – 2ab sin θ cos θ = c² 

a² – a² sin² θ + b² – b² cos² θ – 2ab sin θ cos θ = c² 

a² + b² – c² = a² sin² θ + b² cos² θ + 2ab sin θ cos θ

(i.e.) (a sin θ + b cos θ)² = ±√(a² + b² + c²)

a sin θ + b cos θ = ±√(a² + b² + c²)

The correct answer is (B) 1/√3.

On substituting the values of various T-ratios, we get

= (tan 60° - tan 30°)/(1 + tan 60°. Tan 30°)

= (√3 – 1/√3)/(1 + √3 x 1/√3)

= {(3 – 1)/√3}/(1 + 1)

= 2/2 x √3

= 1/√3

The correct answer is (B) 30°

We have,

2 sin 2 θ = √3

sin 2 θ = √3/2

sin 2 θ = sin 60°

2 θ = 60°

θ= 60°/2

θ = 30°

The correct answer is (B) 3/2

We have,

sin θ = cos θ

sin θ/ cos θ = cos θ/ cos θ  [ Dividing both sides by cos θ]

tan θ = 1

tan θ = tan 45°

θ = 45°

2 tan² θ + sin² θ – 1

= 2 tan² 45° + sin²45° - 1

= 2 (1)² + (1/√2)² – 1

= 2 + 1/2 – 1

= (4 + 1 – 2)/2

= 3/2

The correct answer is (C) 90°

sin A = cos B

cos (90° - A) = cos B   [sin A = cos (90° - A)]

90° - A = B

90° = A + B

A + B = 90°

Correct answer is (C) 1

The common systems of measuring angles are:

(a) Sexagesimal System: Here, one complete revolution is divided into 360 equal parts, each called a degree, a degree is divided into sixty equal parts, each called a second. Thus, 

One complete revolution = 360° 

1° = 60' (minutes) 

1' = 60'' (seconds) 

(b) Circular System: The unit of measuring an angle in this system is radian. A radian is the measure of the central angle subtended by an arc equal in length to the radius of the circle. 1 radian is written as 1^c.

(i) LHS =  sin53° cos37° + cos53°sin37°

= sin(90° - 37°)cos37° + cos(90° - 37°) sin 37°

= cos37°cos37° + sin 37° sin 37°

= cos² 37° + sin² 37°

= 1

= RHS

(ii) LHS =  cos54°cos36° - sin54°sin36°

= cos (90° - 36°)cos36° - sin(90° - 36°) sin 36°

= sin36°cos36° - cos 36° sin 36°

= 0

= RHS

(iii) LHS =  sec70°sin20° + cos20°cosec70°

= sec (90° - 20°)sin20° + cos20°cosec(90°- 20°

= cosec20° . 1/cosec20° + 1/sec 20°. sec20°

= 1 + 1

= 2

= RHS

 2 sin 3x = \(\sqrt3\)

⇒ sin 3x = \(\frac{\sqrt3}2\)

⇒ sin 3x = sin 60°

⇒ 3x = 60°

⇒ x = 20°

Solution
Let QR=x and PR=y
Then x+y=25
y=25-x
Now by Pythagorus theorem
x²  + 25= y²
x²  + 25 = (25-x)²
Solving it ,we get
X=12 cm
Then y=25-12=13 cm
Now sin P= 12/13
cos P=5/13
Tan P=12/5

1. False , the value if tan 60° = √3= 1.73
2. True, The value of sec A increase from 1 to ∞.
3. False, Cos A is the abbreviation used for the cosine of angle A
4. False, cot A is one symbol. We cannot separate it
5. False, The value of sin θ always lies between 0 and 0 and 4/3 > 1

LHS = 2.sec² θ – sec⁴ θ – 2.cosec² θ + cosec⁴ θ =  2 sec² θ – (sec² θ)² – 2cosec² θ + (cosec² θ)²

= 2(1+ tan² θ) – (1+ tan² θ)² – 2(1+ cot² θ) + (1+ cot² θ)²

= 2 + 2tan² θ – (1 + 2tan² θ + tan⁴ θ) – 2 – 2cot² θ + 1 + 2cot² θ + cot⁴ θ 

= 2 + 2.tan² θ – 1 – 2 tan² θ – tan⁴ θ – 2 – 2 cot² θ + 1 + 2 cot² θ + cot⁴ θ 

= cot⁴ θ – tan⁴ θ = R.H.S.

L.H.S. = (tan θ + 1/cos θ)² + (tan θ - 1/cos θ)² 

= (tanθ + secθ)² + (tanθ – secθ)² 

= tan² θ + 2 tan θ sec θ + sec² θ + tan² θ – 2 tan θ sec θ +.sec² θ 

= 2(tan² θ + sec² θ)

L.H.S = tan(π/4+θ)

= \(\frac{tan\frac{\pi}{4}+tan \theta}{1-tan\frac{\pi}{4}tan\theta}\)

= \(\frac{1+tan\theta}{1-(1)tan\theta}\)

=\(\frac{1+tan\theta}{1-tan\theta}\)

=RHS

Correct option is: C) 1

\(\frac{Sin\, \alpha}{Cos\, \alpha} . Cot\, \alpha =\) \(\frac {sin \,\alpha}{cos \, \alpha}. \frac {cos \, \alpha}{sin \, \alpha} = 1\) ( \(\because\) cot \(\alpha\) = \(\frac {cos \, \alpha}{sin \, \alpha}\))

Correct option is: C) 1

The correct option is: (C) 60°, 30°, 0°

Explanation:

We have, sin(A + B + C) = 1

 => sin(A + B + C) = sin 90°

=> A + B + C = 90°    ...(i)

Also, tan(A - B) = 1/√3 = tan 30°

=> A - B = 30°    ...(ii)

and sec (A + C) = 2 = sec 60°

=> A + C = 60°      ....(iii)

From (ii) and (iii), we get

B + C = 30°      ...(iv)

From (i) and (iv), we get, A = 60°

. ^.. B = 30°      [Using A = 60° in (ii)]

and C = 0°      [Using A = 60° in (iii)]

(D) The answer is 1

(b²x² + a²y²)

= b² (a sin θ)² + a² (b cos θ)² 

= b²a² sin² θ + a²b² cos² θ 

= a²b² (sin² θ + cos² θ) 

= a²b² (1) 

= a²b²

Taking LHS = 1 – 2 cos² θ + cos⁴ θ

We know that,

cos² θ + sin² θ = 1

= 1– 2 cos² θ + (cos² θ)²

= 1 – 2 cos² θ + (1 – sin² θ)²

= 1 – 2 cos² θ +1 + sin⁴ θ – 2sin²θ

= 2 – 2(cos² θ + sin²θ) + sin⁴ θ

= 2 – 2(1) + sin⁴ θ

= sin⁴ θ

= RHS

Hence Proved

Taking LHS = 1 – 2 sin² θ + sin⁴ θ

We know that,

cos² θ + sin² θ = 1

= 1– 2 sin² θ + (sin² θ)²

= 1 – 2 sin² θ + (1 – cos² θ)²

= 1 – 2 sin² θ +1 + cos⁴ θ – 2cos²θ

= 2 – 2(cos² θ + sin²θ) + cos⁴ θ

= 2 – 2(1) + cos⁴ θ

= cos⁴ θ

= RHS

Hence Proved

Correct option is: A) \(1-2 \,sin^2\theta\)

\(cos^4 \theta – sin^4 \theta =\) \((cos^2\theta)^2 - (sin^2\theta)^2\)

= \((cos^2\theta - sin^2\theta) (cos^2\theta + sin^2\theta)\)

(\(\because\) \(a^2-b^2 = (1-b) (a+b))\)

= \(cos^2\theta - sin^2\theta \) (\(\because\) \(\cos^2\theta + sin^2\theta =1\))

= \((1-sin^2\theta) - sin^2\theta \) ( \(\because\) \(cos^2\theta = 1-sin^2\theta\) )

= \(1-2 \,sin^2\theta\) 

Correct option is: A) 1 – 2sin²θ

L.H.S. cos⁴θ – sin⁴θ 

= (cos²θ)² – (sin²θ)² 

= (cos²θ + sin²θ) (cos²θ – sin²θ) 

= 1 . (cos²θ – (1 – cos²θ) = cos²θ – 1 + cos²θ 

= 2cos²θ – 1 = R.H.S.

Answer : (c) nπ + α 

Given, 

sin θ = sin α ...(i) 

cos θ = cos α ...(ii) 

∴ Dividing eqn (i) by eqn (ii), we have 

tan θ = tan α 

⇒ θ = nπ + α.

Given, tan 2θ tanθ = 1

⇒ \(\frac{ 2\,tan\theta}{1-tan^2\theta}\) . tanq =1 

⇒ \(\frac{ 2\,tan\theta}{1-tan^2\theta}\) =1 

⇒2 tan² θ= 1− tan² θ

⇒ 3 tan² θ = 1 

⇒ tan² θ = \(\frac{1}{3}\) 

⇒ tan² θ = ( \(\frac{1}{\sqrt{3}}\) )² = tan² \(\frac{π}{6}\) 

⇒ θ = nπ ±  \(\frac{π}{6}\)

(i) 2 sin θ – 1 = 0 

⇒sin θ = \(\frac{1}{2}\) = sin \(\frac{\pi}{6}\) 

⇒ θ = nπ + (–1)ⁿ \(\frac{\pi}{6}\) , n∈I 

(ii) cos θ =  − \(\frac{1}{2}\)

= cos (180° – 60°) = cos ( π - \(\frac{\pi}{3}\) ) = cos \(\frac{2\pi}{3}\)

∴ θ = 2nπ \(\pm\) \(\frac{2\pi}{3}\),  n∈I 

(iii) 4 sin² θ = 1 

⇒ \(\frac{4(1-cos\,2\theta)}{2}\) = 1 

⇒ 2(1 - cos 2θ) =1 ⇒ 1-cos 2θ = \(\frac{1}{2}\) ⇒ cos 2θ = \(\frac{1}{2}\) = cos \(\frac{\pi}{3}\)

∴ 2θ = 2nπ ± \(\frac{\pi}{3}\) 

⇒ θ = nπ ± \(\frac{\pi}{6}\)

(iv) tan 2x – \(\sqrt{3}\) = 0 

⇒ tan 2x = \(\sqrt{3}\) = tan \(\frac{\pi}{3}\) 

⇒ 2x = nπ + \(\frac{\pi}{3}\) 

⇒ x = \(\frac{n\pi}{2}\) + \(\frac{\pi}{6}\) , n ∈ I 

(v) 2 cot² θ = cosec² θ 

⇒ 2 cot² θ = 1 + cot² θ 

⇒ cot² θ = 1

⇒ cot θ = ± 1

= cot (± \(\frac{\pi}{4}\)) 

⇒ θ = nπ ± \(\frac{\pi}{4}\) ,  n ∈ I

Answer : (c) \(\frac{n\pi}{3}\)+ \(\frac{\pi}{12}\)

tan θ + tan 2θ = 1 – tan θ . tan 2θ

⇒ \(\frac{tan\,θ +tan\,2θ}{1- tan\,θ\,tan\,2θ}\) =1

⇒ tan (θ + 2θ) = 1 \(\big( \because tan\,(A+B) = \frac{tan\,A+tan\,B}{1-tan\,A\,tan\,B}\big)\) 

⇒ tan 3θ = tan \(\frac{\pi}{4}\) 

⇒ 3θ = nπ + \(\frac{\pi}{4}\) 

⇒ θ = \(\frac{n\pi}{3}\)+ \(\frac{\pi}{12}\).

Answer : (b) 2kπ, k∈I   

1 – cos θ = sin θ . sin θ/2 

⇒ 2 sin² θ/2 = 2 sin θ/2 . cos θ/2 . sin θ/2 

⇒ 2 sin² θ/2 (1 – cos θ/2) = 0 

⇒ 2 sin² θ/2 = 0 or 1 – cos θ/2 = 0 ⇒ sin θ/2 = 0 or 2 sin² θ/4 = 0 

(Using, 1 – cos 2θ = 2 sin² θ)

⇒ \(\frac{θ}{2}\) = kπ or \(\frac{θ}{2}\) = kπ , where k∈I.

⇒ θ = 2kπ or θ = 4kπ, k∈I. 

⇒ θ = 2kπ, k∈I

Answer : (a)  \(x = nπ ± \frac{π}{4}\)

4 cos² x + 6 sin² x = 5 

⇒ 4 (cos² x + sin² x) + 2 sin² x = 5 

⇒ 2 sin² x = 5 – 4 = 1  ⇒ sin² x = \(\frac{1}{2}\) 

⇒ sin x = ± \(\frac{1}{√2}\) 

⇒ \(x = nπ ± \frac{π}{4}\)

Answer : (c) =  θ = nπ ± \(\big( \frac{\pi}{4} -\frac{\alpha}{2}\big)\)

cos 2θ = sin α ⇒ cos 2θ = cos \(\big( \frac{\pi}{2} -\alpha\big)\) 

⇒ 2θ = 2nπ ± \(\big( \frac{\pi}{2} -\alpha\big)\) (∵ cos θ = cos α ⇒ θ = 2nπ ± α)

⇒ θ = nπ ± \(\big( \frac{\pi}{4} -\frac{\alpha}{2}\big)\)

Correct option is: C) √3

sin \(\theta\) = \(\frac 12\) = sin \(30^\circ\)

= \(\theta\) = \(30^\circ\)

\(\therefore\) cot \(\theta\) = cot \(30^\circ\) = \(\sqrt3\)

Correct option is: C) √3

Given:

sin²x – cos 2x = 2 – sin 2x 

⇒ 1 – cos²x – (2cos²x – 1) 

= 2 – 2 sin x cos x 

⇒ – 3 cos²x + 2 sin x cos x = 0 

⇒ cos x (2 sin x – 3 cos x) = 0 ⇒ cos x = 0 (∵ 2 sin x ≠ 3 cos x)

⇒ x = 2nπ ± \(\frac{\pi}{2}\) 

⇒ x = (4n ± 1) \(\frac{\pi}{2}\) 

Answer : (b) nπ ± \(\frac{π}{6}\),  n ∈ Z 

cos 2x + 2 cos² x = 2 

⇒ 2 cos² x – 1 + 2 cos² x = 2

⇒ 4 cos² x = 3 

⇒ cos x = ± \(\frac{\sqrt{3}}{2}\) 

⇒ x = nπ ± \(\frac{π}{6}\), n∈Z

Answer: (a) 210°, 330° 

Given, cosec θ + 2 = 0 

⇒ cosec θ = –2 

⇒ sin θ = − \(\frac{1}{2}\) = – sin 30° 

⇒ sin θ = – sin 30° = sin (180° + 30°) or sin (360° – 30°) 

⇒ sin θ = sin 210° or sin 330° 

⇒ θ = 210°, 330°.

Answer : (c) \(\frac{π}{12}\)

sin x cos x = \(\frac{1}{4}\) 

⇒ 2 sin x cos x = \(\frac{1}{2}\)  ⇒ sin 2x = \(\frac{1}{2}\) 

⇒ sin 2x = sin \(\frac{π}{6}\) 

⇒ 2x = \(\frac{π}{6}\) , x ∈ ( 0 , \(\frac{π}{2}\) ) 

⇒ x = \(\frac{π}{12}\) 

The solution of a trigonometric equation is a value of the unknown angle that satisfies the equation. 

A trigonometric equation may have an unlimited number of solutions. 

For example, if sin x = 0, then x = 0, π, 2π, 3π, .... 

• A solution lying between 0° and 360° is called the principal solution. 

• Since the trigonometric functions are periodic, a solution generalized by means of periodicity is known as the general solution.

Every equation will have a principal solution as well as a general solution.

Given 

 sin θ = 0 

∵ sin θ = sin 0 = sin π = sin 2π = sin (–2π) = .... = 0 

Therefore sin θ = 0 is satisfied by the following values of θ. 

∴ θ = 0, ± π, ± 2π, ± 3π, ± 4π, ..... 

⇒ The general solution of sin θ = 0 is θ = nπ, n∈I

sin²30° cos²45° - 3 sin² 30° + 2 cos² 90°

= \((\frac{1}{2})^2 (\frac{1}{\sqrt2})^2 \) - 3 \((\frac{1}{2})^2\) + 2 (0)²

= \(\frac{1}{4} \times \frac{1}{2}\) - 3 x \(\frac{1}{4}\) + 0

= \(\frac{1}{8}\) - \(\frac{3}{4}\)

= \(\frac{1-3}{8}\)

= \(\frac{-2}{8}\)

= \(\frac{-1}{4}\)

Given,

cos θ = 0

\(\theta = \pm\, \frac{\pi}{2}\) ,\(\pm\, \frac{3\pi}{2}\) , \(\pm\, \frac{5\pi}{2}\) , .... satisfy the equation cos θ = 0 as 

cos θ = cos (\( \pm\, \frac{\pi}{2}\) ) = cos (\(\pm\, \frac{3\pi}{2}\) ) = ... = 0

⇒ The general solution of cos θ = 0 is \(\theta = (2n+1) \frac{\pi}{2}\) , n ∈ I 

Given,

tan θ = 0

This is satisfied by θ = 0, ± π, ± 2π, .... 

⇒ The general solution of tan θ = 0 is θ = nπ, n ∈ I

With the help of Minimum-Maximum Value Table we can find the Value of sin θ 

Therefore, 

Minimum Value of sin θ = - 1 

and Maximum Value of sin θ = 1

Let log_{cos x} sin x = p. Then, log_{sin x} cos x = \(\frac{1}{p}\). 

∴ log _{cos x} sin x + log _{sin x} cos x = 2 

⇒ p + \(\frac{1}{p}\) = 2 

⇒ p² – 2p + 1 = 0 

⇒ (p – 1)² = 0 

⇒ p = 1 

⇒ log _{cos x} sin x = 1 

⇒ sin x = cos x 

The smallest positive value of x for which sin x = cos x is x = \(\frac{\pi}{4}\) .

sec 2A = cosec(A - 42°)

⇒ cosec (90°- 2A) = cosec(A - 42°)

⇒  90°- 2A = A - 42°

⇒ 3A = 132°

⇒ A = \(\frac{132°}{3} = 44°\)

tan θ = tan α

⇒ \(\frac{sin\,\theta}{cos \,\theta}\) = \(\frac{sin\,\alpha}{cos \,\alpha}\) 

⇒ sin θ cos α - cos θ sin α = 0

⇒ sin (θ - α) = 0 

⇒ θ – α = nπ  

⇒ θ = nπ + α, n∈I.