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 sin(90° - θ) = cos θ

cos (90° - θ) = sinθ

tan (90° - θ) = cotθ

cot (90° - θ) = tanθ

sec(90° - θ) = cosecθ

cosec(90° - θ) = secθ

(i) sin 59° + cos 56°

= sin (90° - 31°) + cos (90° - 34°)

= cos 31° + sin 34°

(ii) tan 65° + cot 49°

= tan (90° - 25°) + cot (90° - 41°)

= cot 25° + tan 41°

(iii) sec 76° + cosec 52°

= sec(90° - 14°) + cosec(90° - 52°)

= cosec 14° + sec 38°

(iv) cos 78° + sec 78°

= cos (90° -78°) + sec(90° - 78°)

 = sin 12° + cosec 12°

L.H.S

= (tan θ + sec θ -1)(tan θ + 1 +sec θ)

= (tan θ + sec θ)² - 1

= tan² θ + sec² θ + 2 tan θ sec θ - 1

= 2 tan² θ + 2 tan θ sec θ         [ sec² θ  - 1 = tan² θ ]

= 2 tan θ  ( tan θ + sec θ)

= 2 sin θ / cos θ  ( (sin θ / cos θ)  + (1/ cos θ) )    [tan θ = sin θ / cos θ , sec θ  = 1/cos θ ] 

= 2 sin θ / cos θ ( (1 + sin θ)/ cos θ)

= 2 sin θ (1 + sin θ)/ cos² θ

= 2 sin θ (1 + sin θ)/ ( 1- sin² θ)    [ cos² θ = 1 - sin² θ]

= 2 sin θ (1 + sin θ)/(1 - sin θ) (1 + sin θ)

= 2 sin θ / (1 - sin θ) = R.H.S

L.H.S = R.H.S

Correct option is: A) 1

\(Sin^6 A + cos^6 A + 3\, sin^2 A \, cos^2 A\)

= \((Sin^2A + cos^2A)^3 - 3\, sin^2 A \, cos^2 A (Sin^2 A + cos^2 A) + 3\, sin^2A \, cos^2A\)

(\(\because\) \(a^6 + b^6 = (a^2 + b^2 )^3 - 3 \, a^2 b^2 (a^2+b^2)\))

= \(1^3 - 3\, sin^2A \, cos^2A + 3\, sin^2A\, cos^2A\)

(\(\because\) \(sin^2A + cos^2A = 1\))

= 1

Correct option is: A) 1

L.H.S. = sin⁶ A + cos⁶ A 

= (sin² A)³ + (cos² A)³ 

= (sin² A + cos² A)³ – 3sin² A cos² A(sin² A + cos² A) 

…[ a³ + b³ = (a + b)³ – 3ab(a + b)] 

= 1³ – 3sin² A cos² A (1) 

= 1 – 3sin² A cos² A 

= 1 – 3 sin² A (1 – sin² A) 

= 1 – 3 sin² A + 3sin⁴ A 

= R.H.S.

3sin10° – 4sin³10° is of the form 3 sin A – 4sin³A 

= sin3A = sin 3- 10° 

= sin 30° 

= \(\frac{1}{2}\)

1856° = 5 x 360° + 56° 

∴ 1856° and 56° are co-terminal angles. 

Since 0° < 56° < 90°, 56° lies in the 1st quadrant. 

∴ 1856° lies in the 1st quadrant, 

∴ sin 1856° >0 …(i) 

2006° = 5 x 360° + 206° 

∴ 2006° and 206° are co-terminal angles. 

Since 180° < 206° < 270°, 206° lies in the 3rd quadrant. 

∴ 2006° lies in the 3rd quadrant, 

∴ sin 2006° <0 …(ii) 

From (i) and (ii), 

sin 1856° is greater.

L.H.S. = sin⁸ θ – cos⁸ θ 

= (sin⁴ θ)² – (cos⁴ θ)² 

= (sin⁴ θ – cos⁴ θ) (sin⁴ θ + cos⁴ θ) 

= [(sin² θ)² – (cos² θ)² ] 

. [(sin² θ)² + (cos² θ)² ] 

= (sin² θ + cos² θ) (sin² θ – cos² θ). [(sin² θ + cos² θ)² – 2sin² θ.cos² θ] 

…[Y a² + b² = (a + b)² – 2ab] 

= (1) (sin² θ – cos² θ) (1² – 2sin² θ cos² θ) 

= (sin² θ – cos² θ) (1 – 2sin² θ cos² θ) 

= R.H.S.

Since 270° <310° <360°, 

310° lies in the 4th quadrant. 

∴ sin (310°) < 0 

-310° = -360°+ 50° 

∴ 50° and – 310° are co-terminal angles. 

Since 0° < 50° < 90°, 50° lies in the 1st quadrant

∴ – 310° lies in the 1st quadrant. 

∴ sin (- 310°) > 0 

∴ sin (- 310°) is positive.

1 – 2 sin θ cos θ 

= sin² θ + cos² θ – 2sin θ cos θ 

= (sin θ – cos θ)² ≥ 0 for all θ ∈ R

L.H.S. = (sinθ + cosecθ)² + (cos θ + see θ)² 

= sin² θ + cosec² θ + 2sinθ cosec θ + cos² θ + sec² θ + 2sec0 cos0 

= (sin² θ + cos² θ) + cosec² θ + 2 + sec² θ + 2 

= 1 + (1 + cot² θ) + 2 + (1 + tan² θ) + 2 = tan² θ + cot² θ + 7 

= R.H.S

We have

cosB = 3/5 

⇒ cos(90° − A) = 3/5 

(As, A + B = 90° ) 

∴ sinA= 3/5

We have

cotA = 4/3 

⇒ cot(90° − B) = 4/3  (As, A + B = 90° ) 

∴ tanB = 4/3

Given, 

3 tan θ = 4 

⇒ tan θ = \(\frac{4}{3}\)

\(\frac{4 cos\, θ\, -\, sin\, θ}{2 cos\, θ\, +\, sin\, θ}\)

From, let’s divide the numerator and denominator by cos θ. We get,

\(\frac{(4\, –\, tan\, θ)}{(2\, +\, tan\, θ)}\) 

⇒ \(\frac{(4 – (4/3))}{(2 + (4/3))}\) [using the value of tan θ] 

⇒ \(\frac{(12 – 4)}{(6 + 4)}\) [After taking LCM and cancelling it] 

⇒ \(\frac{8}{10}\)

= \(\frac{4}{5}\)

\(\frac{4 cos\, θ\, -\, sin\, θ}{2 cos\, θ\, +\, sin\, θ}\) = \(\frac{4}{5}\)

LHS= (cos²θ-sinθcosecθ+sin²θ)/sinθ

= (cos²θ+sin²θ-1)/sinθ  [Therefore sinθcosecθ=1]

=0

=LHS =RHS Hence proved

Given, 

3 cot θ = 2 

⇒ cot θ = \(\frac{2}{3}\)

\(\frac{4sin\, θ\, -\, 3cos\, θ}{2sin\,θ \,+\, 6 cos\, θ}\)

From, let’s divide the numerator and denominator by sin θ. We get,

\(\frac{(4 –3 cot θ) }{(2 + 6 cot θ)}\)

⇒ \(\frac{(4 – 3(2/3)) }{(2 + 6(2/3))}\) [using the value of tan θ] 

⇒ \(\frac{(4 – 2)}{(2 + 4)}\) [After taking LCM and simplifying it] 

⇒\(\frac{2}{6}\) 

= \(\frac{1}{3}\)

\(\frac{4sin\, θ\, -\, 3cos\, θ}{2sin\,θ \,+\, 6 cos\, θ}\) = \(\frac{1}{3}\).

Given, tan θ = \(\frac{a}{b}\)

\(\frac {a sin\, θ - b cos\, θ}{a sin\, θ + b cos\, θ}\)

From LHS, let’s divide the numerator and denominator by cos θ. 

And we get, 

\(\frac{(a\, tan\, θ\, –\, b)}{(a\, tan\, θ\, +\, b\,) }\)

⇒ \(\frac{(a(a/b) – b)}{(a(a/b) + b)}\) [using the value of tan θ] 

⇒ \(\frac{(a^2 – b^2)}{b^2} \over \frac{(a^2 + b^2)}{b^2}\) [After taking LCM and simplifying it] 

⇒ \(\frac{(a^2 – b^2)}{(a^2 + b^2)}\) = RHS

Hence Proved

Correct option is: B) 2

We have Tan \(\theta\) + Cot \(\theta\) = 2

= \((tan \, \theta + cot \, \theta)^2 = 2^2 = 4\)

= \(tan^2\theta + cot^2\theta + 2 \,tan \, \theta \, cot \, \theta = 4 \) (\(\because\) \((a+b)^2 = a^2+b^2 + 2ab\))

= \(tan^2\theta + cot^2\theta + 2 = 4\) (\(\because\) cot \(\theta\)  \(\frac 1{tan \, \theta} =\) tan \(\theta\) cot \(\theta\) =1)

= \(tan^2\theta + cot^2\theta = 4-2 =2\)

Correct option is: B) 2

Correct option is: C) 1

Sin \(\theta\) . Cot \(\theta\) . Sec \(\theta\) = Sin \(\theta\) . \(\frac {cos\, \theta}{sin \, \theta}. \frac 1 {cos \, \theta }\)

(\(\because\) cot \(\theta\) = \(\frac {cos\, \theta}{sin\, \theta}\) & sec \(\theta\) = \(\frac 1 {cos \, \theta }\))

= 1

Correct option is: C) 1

As we know,

\(\frac{1}{cosecθ}\) = sinθ

And,

Maximum value of sinθ = 1

so,

The maximum value of \(\frac{1}{cosecθ}\) is 1.

Correct answer is (b) 17/4

We have (cos θ + sec θ) = 5/2 

Squaring both sides, we get: 

(Cos θ + sec θ)² = (5/2)² 

⇒ cos² θ + sec² θ + 2θ = 25/4 

⇒ cos² θ + sec² θ + 2 = 25/4 [∵ sec θ = 1/cos θ] 

⇒ cos² θ + sec² θ = 25/4 − 2 = 17/4

Correct answer is (B) 0

Answer : (a) θ = \(\frac{nπ}{7}\) + \(\frac{π}{14}\)

tan 5θ = cot 2θ 

⇒ tan 5θ = tan (π/2 – 2θ) 

⇒ 5θ = nπ + (π/2 – 2θ), n∈I 

⇒ 7θ = nπ + π/2, n∈I 

⇒ θ = \(\frac{nπ}{7}\) + \(\frac{π}{14}\) , n∈I.

We know that, cos (90° − θ) = sin θ. 

So, 

cos 48° – sin 42°

= cos (90° − 42°) – sin 42°

= sin 42° – sin 42°

= 0 

Thus the value of cos 48° – sin 42° is 0.

(a) 10π seconds

1 rotation makes 2π 

Distance travelled in 1 second = 2 radians 

So time taken to complete 10 rotations = 10 × 2π = 20 π^c 

= 20π/2 = 10π seconds

LHS = a cos A+ 6 cos B + c cos C 

Using sine formula, we get k sin A cos A + k sin B cos B + k sin C cos C k 

= (k/2) [2 sin A cos A + 2 sin B cos B + 2 sin C cos C] 

= (k/2) [sin 2A + sin 2B + sin 2C] 

= (k/2) [2 sin (A + B) . cos (A – B) + 2 sin C . cos C] 

= (k/2) [2 sin (A – B) . cos (A – B) + 2 sin C . cos C] 

= (k/2) [2 sin C . cos (A – B) + 2 sin C . cos C] 

= k sin C [cos(A – B) + cos C]

= k sin C[cos(A - B) + cos(π - bar(A + B))

= k sin C [cos (A – B) – cos (A + B)] 

= k sin C . 2 sin A sin B

= 2k sin A . sin B sin C 

= 2a sin B sin C = RHS

 Correct option is: A) \(\frac{p^2-1}{p^2+1}\)

We have 

sec \(\theta\) + tan \(\theta\) = P ....(1)

On multiplying both sides of equation (1) by(sec \(\theta\) - tan \(\theta\) ), we get

(sec \(\theta\) + tan \(\theta\) ) (sec \(\theta\) - tan \(\theta\) ) = P ( sec \(\theta\) - tan \(\theta\) )

= P ( sec \(\theta\) - tan \(\theta\)) = \(\sec^2\, \theta - tan^2 \, \theta = 1\)  (\(\because\) \(\sec^2\, \theta - tan^2 \, \theta = 1\))

= sec \(\theta\) - tan \(\theta\) = \(\frac 1P\) ....(2)

By adding equations (1) & (2), we get

(sec \(\theta\) + tan \(\theta\)) + ( sec \(\theta\) - tan \(\theta\)) = P + \(\frac 1P\)

= 2 sec \(\theta\) = \(\frac {P^2+1}{P}\) 

= sec \(\theta\) = \(\frac {P^2+1}{2P}\) ....(3)

By subtracting equation (2) from equation (1), we get

(sec \(\theta\) + tan \(\theta\)) - ( sec \(\theta\) - tan \(\theta\)) = P - \(\frac 1P\)

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

= tan \(\theta\) = \(\frac {P^2-1}{2P}\) ....(4)

Now, sin \(\theta\) = \(\frac {\frac {sin\, \theta}{cos\, \theta}}{\frac 1{cos\, \theta}} = \frac {tan\, \theta}{sec\, \theta}= \frac {\frac {P^2-1}{2P}}{\frac {P^2+1}{2P}} = \frac {P^2-1}{P^2+1}\)  (From (3) & (4)

Correct option is: A) \(\frac{p^2-1}{p^2+1}\)

sec θ = √2 ⇒ cos θ = \(\frac1{\sqrt2}\)

∴ sin θ = ±\(\sqrt{1-cos^2\theta}\) = ±\(\sqrt{1-\frac12}\) = ±\(\frac1{\sqrt2}\)

Since θ lies in the fourth quadrant, so sin θ is –ve and cos q is +ve.

∴ sin θ = \(-\frac1{\sqrt2}\), cosec θ = √2

tan θ = \(\frac{sin\,\theta}{cos\,\theta}\) = \(-\frac1{\sqrt2}\) x \(\frac{\sqrt2}{1}\) = –1 ⇒ cot θ = –1

∴ \(\frac{1+tan\,\theta+cosec\,\theta}{1+cot\,\theta-cosec\,\theta}\) = \(\frac{1-1-\sqrt2}{1-1+\sqrt2}= -1.\)

Answer is (A)

LHS = sin 2A = sin0° = 0 

RHS = 2sin A = 2sin0° = 0

Taking LHS = tan⁴ θ + tan² θ

= (tan² θ)² + tan² θ

= ( sec² θ – 1)² + (sec² θ – 1) [∵ 1+ tan² θ = sec² θ ]

= sec⁴ θ + 1 – 2 sec² θ + sec² θ – 1 [∵ (a – b)² = (a² + b² – 2ab)]

= sec⁴ θ – sec² θ

= RHS

Hence Proved

(C) 1

According to the question,

cos 9∝ = sin ∝ and 9∝<90°

i.e. 9α is an acute angle

We know that,

sin(90°-θ) = cos θ

So,

cos 9∝ = sin (90°-∝)

Since, cos 9∝ = sin(90°-9∝) and sin(90°-∝) = sin∝

Thus, sin (90°-9∝) = sin∝

90°-9∝ =∝

10∝ = 90°

∝ = 9°

Substituting ∝ = 9° in tan 5∝, we get,

tan 5∝ = tan (5×9) = tan 45° = 1

∴, tan 5∝ = 1

The ratio of the length of rod and its shadow = 1: √3 

Let the angle of elevation of sun be θ 

tan θ = \(\frac{P}{B}\) (P = perpendicular, B = base) 

Here tan θ = 1: √3 = \(\frac{1}{\sqrt3}\)

tan θ = \(\frac{\sqrt3}{3}\) (by rationalizing the denominator) 

θ = 30° (∵ tan 30° = (\(\frac{\sqrt3}{3}\))

Correct answer is (a) 1

cos A + cos² A = 1 

=> cos A = 1 − cos² A 

=> cos A = sin² A (∵ 1 − cos² A = sin²) 

=> cos² A = sin⁴ A (Squaring both sides) 

=> 1 − sin² A = sin⁴ A 

=> sin⁴ A + sin² A = 1

Taking LHS = cos⁴ A + sin⁴ A + 2 sin² A cos² A

Using the identity,(a + b)² = (a² + b² + 2ab)

Here, a = cos² A and b = sin² A

= ( cos² A + sin² A) [∵ cos² θ + sin² θ = 1]

= 1

A. 1

Given: sin A + sin² A = 1

⇒ sin A = 1 – sin² A

(By Rearranging)

⇒ sin A = cos² A

(∵, sin² θ +cos² θ = 1)

⇒ cos² θ = 1 – sin² θ)

Squaring both sides, we get

⇒ sin² A = cos⁴ A or cos⁴ A = sin² A

Thus, cos² A + cos⁴ A = sin A + sin² A = 1 (∵ sin A+sin² A = 1)

∴ cos² A + cos⁴ A = 1

Correct answer is (b) 1

Sin A + sin² A = 1 

=> sin A = 1 − sin² A 

=> sin A = cos² A (∵ 1 − sin² A) 

=> sin² A = cos⁴ A (Squaring both sides) 

=> 1 − cos² A = cos⁴ A 

=> cos⁴ A + cos² A = 1

Correct option is: A) \( \frac{3}{5}\)

We have cosec \(\theta\) + cot \(\theta\) = 2

= \(\frac {1}{sin \theta} + \frac {cos \theta}{sin \theta} =2\)

= 1 + cos \(\theta\) = 2 sin \(\theta\)

= \((1 + cos \theta )^2 = 4\, sin^2\theta\) (By squaring both sides)

= 1 + \(cos^2\theta + 2\, cos \theta = 4 (1- cos^2\theta)\) (\(\because\) \(sin^2\theta = 1- cos^2\theta\))

= \(5 \,cos ^2\theta + 2\, cos \theta - 3 = 0\)

= \(5 \,cos ^2\theta + 5 \,cos\theta - 3 \, cos\theta - 3 = 0\)

= \(5 \,cos\theta (cos\theta +1) - 3 (cos\theta +1 ) = 0\)

= (5 cos \(\theta\) - 3 ) (cos \(\theta\) + 1) = 0

= 5 cos \(\theta\) - 3 = 0 or cos \(\theta\) + 1 = 0

= cos \(\theta\) = \(\frac 35\) or cos \(\theta\) = -1

cos \(\theta\) \(\neq\) -1

\(\because\) If cos \(\theta\) = -1 then sin \(\theta\) = \(\sqrt {1-cos^2\theta} = \sqrt {1-(-1)^2} = \sqrt {1-1} = 0\)

Then cosec \(\theta\) = \(\frac 1{sin \theta} = \frac 10 = \infty\) (Not defined)

\(\therefore\) cos \(\theta\) = \(\frac 35\)

Correct option is: A) \(\frac{3}{5}\)

Correct option is: A) \(\frac{1}{2}\)

Given that \(4\, cos^2\theta -3 = 0\)

\(\therefore\) \(cos^2\theta = \frac 34\)

 \(1- sin^2\theta = \frac 34\) (\(\because\) \(cos^2\theta = 1-sin^2\theta)\)

\(Sin^2\theta = 1 - \frac 34 = \frac 14\) 

sin \(\theta\) = \(\pm \frac 12\)

sin \(\theta\)  = \(\frac 12\) or sin \(\theta\) = \(\frac {-1}2\)

Correct option is: A) \(\frac{1}{2}\)