Explore topic-wise InterviewSolutions in .

False

Justification:

We know that,

sin θ increases when 0° ≤ θ ≤ 90°

cos θ decreases when 0° ≤ θ ≤ 90°

And (sin 80°-cos 80°) = (increasing value-decreasing value)

= a positive value.

Therefore, (sin 80°-cos 80°) > 0.

Correct answer = (c) 1

We have

(sin 43° cos 47° + cos 43° sin 47°)

= sin 43° cos(90° − 43°) + cos 43° sin(90° − 43°) 

= sin 43° sin 43° + cos 43° cos 43° [∵ cos(90° − θ) = sin θ and sin(90° − θ) = cos θ] 

= sin² 43° + cos² 43°

= 1

True

We know that sinθ increases as θ increases but cosθ decreases as θ increases. 

We have tan θ = sin θ/cos θ 

Now as θ increases, sinθ increases but cosθ decreases. Therefore, in case of tanθ, the numerator increases and the denominator decreases. But in case of sinθ which can be seen as sin θ/1, only the numerator increases but the denominator remains fixed at 1. 

Hence tanθ increases faster than sinθ as θ increases.

False

Justification:

Since, (a²-b²) = (a+b)(a-b)

cos² 23° – sin² 67° =(cos 23°+sin 67°)(cos 23°-sin 67°)

= [cos 23°+sin(90°-23°)] [cos 23°-sin(90°-23°)]

= (cos 23°+cos 23°)(cos 23°-cos 23°) (∵sin(90°-θ) = cos θ)

= (cos 23°+cos 23°).0

= 0, which is neither positive nor negative

3 sin θ + 4 cos θ = 5 

⇒ (3 sin θ + 4 cos θ)² = 25 ⇒ 9 sin² θ + 16 cos² θ + 24 sin θ cos θ = 25 

⇒ 9(1 – cos² θ) + 16 (1 – sin² θ) + 24 sin θ cos θ = 25     (∵ sin² θ + cos² θ = 1) 

⇒ 9 – 9 cos² θ + 16 – 16 sin² θ + 24 sin θ cos θ = 25 

⇒ 9 cos² θ + 16 sin² θ – 24 sin θ cos θ = 0 

⇒ (3 cos θ – 4 sin θ)² = 0 ⇒ 3 cos θ – 4 sin θ = 0.

i. sin 20° = cos (90° – 20°) …..[∵ sin θ = cos (90 – θ)] 

= cos 70° 

ii. tan θ x tan (90 – θ) = 1 

Substituting θ = 30°,

 tan 30° x tan (90 – 30)° = 1 

∴ tan 30° x tan 60° = 1 

iii. cos 40° = sin (90° – 40°) …[∵ COS θ = sin (90 – θ)] 

= sin 50°

Rays of sunlight are parallel. 

So, ∆PQR and ∆ABC are equiangular i.e., similar triangles.

Sides of similar triangles are proportional. 

∴ \(\frac{OR}{BC}\) = \(\frac{PR}{AC}\)

∴ Height of the tree (QR) = \(\frac{BC}{AC}\) x PR 

Substituting the values of PR, BC and AC in the above equation, we can get length of QR i.e., the height of the tree.

At 8’O clock in the morning, the sunlight is not very bright. At the same time, the sun is on the horizon and the shadow would by very long. It would be extremely difficult to measure shadow in that case. Between 11:30 am and 1:30 pm, the sun is overhead and it would be easier to measure the length of shadow.

i. sin P = \(\cfrac{Opposite \,of ∠ P}{Hypotenuse}\) = \(\frac{QR}{PQ}\)

ii .cos Q = \(\cfrac{Adjacent\, side \,of ∠ Q}{Hypotenuse}\) = \(\frac{QR}{PQ}\)

iii. tan P = \(\cfrac{Opposite \,side\,of ∠ P}{Adjacent\, side \, of\,∠P}\) = \(\frac{QR}{PR}\)

 iv. tan Q = \(\cfrac{Opposite \,side\,of ∠ Q}{Adjacent\, side \, of\,∠Q}\) = \(\frac{PR}{QR}\)

 i. sin X = \(\cfrac{Opposite \,side \,of ∠ X}{Hypotenuse}\) = \(\frac{YZ}{XZ}\) = \(\frac{a}{c}\)

 ii. tan Z = \(\cfrac{Opposite \,side \,of ∠ Z}{Adjaent \,side \,of \, ∠ Z}\) = \(\frac{XY}{YZ}\) = \(\frac{b}{a}\)

  iii. cos X = \(\cfrac{Adjacent \,side \,of ∠ X}{Hypotenuse}\) = \(\frac{XY}{XZ}\) = \(\frac{b}{c}\) 

 ii. tan X = \(\cfrac{Opposite \,side \,of ∠ X}{Adjaent \,side \,of \, ∠ X}\) = \(\frac{XZ}{XY}\) = \(\frac{a}{b}\)

 i. sin 50° = \(\cfrac{Opposite \,side \,of \,50°}{Hypotenuse}\) = \(\frac{MN}{LN}\) 

ii.cos 50° = \(\cfrac{Adjacent \,side \,of \,50°}{Hypotenuse}\) = \(\frac{LM}{LN}\) 

iii. tan 40° = \(\cfrac{Opposite\,side \,of \,40°}{Adjacent\,side\, of \,40°}\) = \(\frac{LM}{MN}\) 

iv. cos 40° = \(\cfrac{Adjacent \,side \,of \,40°}{Hypotenuse}\) = \(\frac{MN}{LN}\) 

LHS = sin (50° + θ) – cos (40° - θ)

We know that,

Sin A = cos (90° - A)

Here, A = 50° + θ

⇒ cos {90° -( 50° + θ)} – cos (40° - θ)

⇒ cos (40° - θ) – cos (40° - θ)

= 0 = RHS

Hence Proved

 cos2B = \(\frac{cos\,(A+B)}{cos\,(A-C)}\)

⇒ \(\frac{1-tan^2\,B}{1+tan^2\,B}\) = \(\frac{cos\,A\,cos\,C-sin\,A\,sin\,C}{cos\,A\,cos\,C\,sin\,A\,sin\,C}\)

⇒ \(\frac{1-tan^2\,B}{1+tan^2\,B}\) = \(\frac{1-tan\,A\,tan\,C}{1+tan\,A\,tan\,C}\)             (On dividing the numerator and denominator of RHS by cos A cos C)

⇒ 1 + tan A tan C – tan² B – tan A tan² B tan C = 1 + tan² B – tan A tan C – tan A tan² B tan C 

⇒ 2 tan A tan C = 2 tan² B ⇒ tan A . tan C = tan² B ⇒ tan A, tan B, tan C are in G.P

3cos θ + 4sin θ = 4 

∴ 3cos θ = 4(1 – sin θ) 

Squaring both the sides, we get . 

9cos² θ = 16(1 – sin θ)² 

∴ 9(1 – sin² θ) = 16(1 + sin² θ – 2sin θ) 

∴ 9 – 9sin² θ = 16 + 16sin² θ – 32sin θ 

∴ 25sin² θ – 32sin θ + 7 = 0 

∴ 25sin² θ – 25sin θ – 7sin θ + 7 = 0 

25sin θ (sin θ – 1) – 7 (sin θ – 1) = 0 

∴ (sin θ – 1) (25sin θ – 7) = 0 

∴ sin θ – 1 = 0 or 25 sin θ – 7 = 0 

∴ sin θ = 1 or sin θ = 7/25

Since, -1 ≤ sin θ ≤ 1 

∴ sin θ = 1 or 7/25

(b) \(\frac34\)

4 cot² 45° – sec² 60° + sin² 60° + cos² 90° 

= 4 (cot 45°)² – (sec 60°)² + (sin 60°)² + (cos 90°)²

= 4 × 1² – (2)² + \(\bigg(\frac{\sqrt3}{2}\bigg)\) + (0)²

= 4 - 4 + \(\frac34\) + 0 = \(\frac34\).

Correct option is: A) Sec θ + Tan θ

\(\frac{1+Sin \theta}{\sqrt{1-Sin^2 \theta}} =\) \(\frac{1+Sin \theta}{ \sqrt{cos^2 \theta}} = \frac {1+sin\, \theta}{cos\, \theta}\) 

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

= sec \(\theta\) + tan \(\theta\)

Correct option is: A) Sec θ + Tan θ

L.H.S: (√3 + 1) (3 – cot 30°)

= (√3 + 1) (3 – √3) [∵cos 30° = √3]

= (√3 + 1) √3 (√3 – 1) [∵(3 – √3) = √3 (√3 – 1)]

= ((√3)²– 1) √3 [∵ (√3+1)(√3-1) = ((√3)² – 1)]

= (3-1) √3

= 2√3

Similarly solving R.H.S: tan³ 60° – 2 sin 60°

Since, tan 60^o = √3 and sin 60^o = √3/2,

We get,

(√3)³ – 2.(√3/2) = 3√3 – √3

= 2√3

Therefore, L.H.S = R.H.S

Hence, proved.

Answer is (2) 7

(sin α + cosec α)² + (cos α + sec α)² = k + tan² α + cot² α

sin² α + cosec² α + 2 sin α cosec α + cos² α + sec² α + 2 cos α sec α = k + tan² α + cot² α

sin² α + cos² α + cosec² α + sec² α + 2 sin α x + \(\frac{1}{sin\alpha}\) 2 cos α x \(\frac{1}{cos\alpha}\) = k + tan² α + cot² α

1 + 1 + cot² α + 1 + tan² α + 2 + 2 = k + tan² α + cot² α

7 + cot² α + tan² α = k + tan² α + cot² α

∴ k = 7

(2) b² – a²

p² – q² = (p + q) (p – q)

= (a cot θ + b cosec θ + b cot θ + a cosec θ) (a cot θ + b cosec θ – b cot θ – a cosec θ)

= [cot θ (a + b) + cosec θ (a + b)] [cot θ (a – b) + cosec θ (b – a)]

= (a + b) [cot θ + cosec θ] (a – b) [cosec θ (a – b)]

= (a + b) [cot θ + cosec θ] (a – b) [cot θ – cosec θ]

= (a + b) (a – b) (cot² θ – cosec² θ)

= (a² – b²) (-1) = – (a² – b²)

p² – q² = b² – a²

(b) 1 – sin² θ cos² θ     

\(\frac{sin^6\,θ-cos^6\,θ}{sin^2\,θ-cos^2\,θ}\) = \(\frac{(sin^2\,\theta)^3-(cos^2\,\theta)^3}{sin^2\,\theta-cos^2\,\theta}\)

= \(\frac{(sin^\,\theta-cos^2\,\theta)(sin^4\,\theta+cos^4\,\theta+sin^2\,\theta\,\cos^2\,\theta)}{(sin^2\,\theta-cos^2\,\theta)}\)         (Using a³ – b³ = (a – b) (a² + b² + ab)) 

= sin⁴ θ + cos⁴ θ + 2 sin² θ cos² θ – sin² θ cos² θ 

= (sin² θ + cos² θ)² – sin² θ cos² θ 

= 1 – sin² θ cos² θ                       (∵ sin² θ + cos² θ = 1)

(c) \(\fracπ4\)

\(\frac{cos\,x}{1+cosec\,x}+\frac{cos\,x}{cosec\,x-1}=2\)

⇒ \(\frac{cos\,x(cosec\,x-1)+cos\,x(1+cosec\,x)}{cosec^2\,x-1}=2\)

⇒ \(\frac{2\,cos\,x\,cosec\,x}{cot^2\,x}=2\)

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

⇒ tan \(x\) = 1 ⇒ tan\(x\) = tan \(\fracπ4\) ⇒ \(x\) = \(\fracπ4\).

(b) cos 2θ

cos 2 (θ + ϕ) + 4 cos (θ + ϕ) sin θ sin ϕ + 2sin²ϕ 

= {cos 2θ cos 2ϕ – sin 2θ sin 2ϕ} + 4 {(cos θ cos ϕ – sin θ sin ϕ) sin θ sin ϕ} + 2 sin²ϕ 

= {(1 – 2 sin² θ) (1 – 2 sin²ϕ) – 2 sin θ cos θ . 2 sin ϕ cos ϕ} + [4 cos θ cos ϕ sin θ sin ϕ – 4 sin²θ sin²ϕ] + 2 sin²ϕ 

= 1 – 2 sin² θ – 2 sin²ϕ + 4 sin²θ sin2 ϕ – 4 sin θ sin ϕ cos θ cos ϕ + 4 cos θ cos ϕ sin θ sin φ – 4 sin²θ sin²ϕ + 2 sin²ϕ 

= 1 – 2 sin²θ = cos 2θ.

The correct option is: (A) x² + y² + z² = r²

Explanation:

x = r sinθ cos ϕ       ...(i)

y = r sinθ sin ϕ        ....(ii)

z = r cos θ               ....(iii)

Squaring and adding (i) and (ii), we get

 x² + y²  = r² sin²θ      ....(iv)

Squaring (iii) and adding it with (iv), we get

x² + y² + z² = r²

(a)  \(\bigg(\frac{x-a}{a}\bigg)^2\) +  \(\bigg(\frac{y-b}{b}\bigg)^2\) + \(\bigg(\frac{z-c}{c}\bigg)^2\)

Given, \(x\) = a (1 + cos θ cos ϕ) 

⇒ \(\frac{x}{a}\) = 1+ cos θ cos  ϕ ⇒ \(\frac{x}{a}\) - 1 cos θ cos ϕ 

⇒ \(\frac{x-a}{a}\) = cos θ cos ϕ                ...(i) 

Similarly, y = b (1 + cos θ sin ϕ ) 

⇒ \(\frac{y-b}{b}\) = cos θ sin ϕ                ...(ii) 

z = c (1+sin θ) ⇒ \(\frac{z-c}{c}\) = sin θ          ...(iii) 

Squaring eqns (i), (ii) and (iii) and adding, we get

\(\bigg(\frac{x-a}{a}\bigg)^2\) +  \(\bigg(\frac{y-b}{b}\bigg)^2\) + \(\bigg(\frac{z-c}{c}\bigg)^2\)

= cos² θ cos² ϕ + cos² θ sin² ϕ + sin² θ 

= cos² θ (cos² ϕ + sin² ϕ) + sin² θ 

= cos² θ + sin² θ = 1.

(i) LHS = \(\frac{sin70°}{cos20°}+\frac{cosec20°}{sec70°}-2cos70°cosec20°\)

= \(\frac{sin70°}{sin(90° - 20°)}+\frac{sec(90°-20°)}{sec70°}- 2cos70°sec(90°-20°)\)

= \(\frac{sin70°}{sin70°}+\frac{sec70°}{sec70°}-2cos70°sec70°\)

= 1 + 1 - 2 x cos70° x\(\frac{1}{cos70°}\)

= 2 - 2

= 0

 = RHs

(ii) LHS = \(\frac{cos80°}{sin10°}+cos59°cosec31°\)

= \(\frac{cos80​°}{sin(90°-10°)}+sin(90°-59°)cosec31°\)

= \(\frac{cos80°}{cos80°}+sin31°cosec31°\)

= \(\frac{cos80°}{sin10°}+cos59°cosec31°\)

= 1 + sin31° x \(\frac{1}{sin31°}\)= 1 + 1

= 2

= RHS

(iii) LHS = \(\frac{sin18°}{sin72°}+\sqrt3(tan10°tan30°tan40°tan50°tan80°) \)

= \(\frac{sin18°}{sin90°-72°}+\sqrt3[cot(90°-10°)\times\frac{1}{\sqrt3}\times{cot(90°-40°)}\times{tan50°}\times{tan80°}]\)

= \(\frac{sin18°}{sin18°}+ \sqrt3(\frac{cos80°\times{cot50°}\times{tan50°}\times{tan80°})}{\sqrt3}\)

= 1 + 1

= 2

= RHS

(i) LHS =  sin35° sin55° - cos35°cos55°

= sin35°cos(90° - 55) - cos35°sin(90° - 55°)

= sin35°cos35° - cos 35° sin 35°

= 0

= RHS

(ii) LHS =  (sin72° + cos18°)(sin72° - scos18°)

= (sin72° + cos18°)(cos(90° - 72°)- cos 18°)

= (sin72° + cos18°) (cos 18° - cos18°)

= (sin72° + cos18°)(0)

= RHS

(iii) LHS = tan48° tan23° tan30° tan42° tan67°

= cot (90-48°) cot (90-23°) tan42° tan 67°

= cot42° cot67° tan 42° tan 67°

= \(\frac{1}{tan 42^\circ} \times \frac{1}{tan 67 ^\circ} \times tan 42^\circ \times tan 67^\circ\)

= 1

= RHS

(i) LHS = tan² 66° - cot² 24°

=  tan²(90° - 24°) - cot² 24°

= cot² 24° - cot² 24°

= 0

= RHS

(ii) LHS =  sin² 48° + sin² 42°

= sin²(90° - 42°) + sin² 42°

= cos² 42° + sin² 42°

= 1

= RHS

(iii) LHS =  cos² 57° - sin² 33°

= cos² (90° - 33°) + sin² 33°

= sin² 33° - sin² 33°

= 0

= RHS

(iv) LHS = (sin 65° + cos25°) (sin65° - cos 25°)

= sin² 65° - cos² 25°

= sin²(90° - 25°) - cos² 25°

= cos² 25° - cos² 25°

= 0

= RHS

Given: sin (A + B) = sin A cos B + cos A sin B and

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

To find: sin 75°

Put A = 30° and B = 45°, then

sin 75° = sin 30° cos 45° + cos 30° sin 45°

= (1/2) × (1/√2) + (√3/2) × (1/√2)

= (1/2√2) + (√3/2√2)

= (1+√3)/2√2

Given: sin (A + B) = sin A cos B + cos A sin B and

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

Find cos 75°

Put A = 45° and B = 30°, then

cos 15° = cos 45° cos 30° + sin 45° sin 30°

= (1/√2) × (√3/2) + (1/√2) × (1/2)

= (√3 / 2√2) + (1/2√2)

= (1+√3)/2√2

(x, y) = (5, 5) 

∴ r = \(\sqrt{x^2+y^2} = \sqrt{25+25}\)

= \(\sqrt{50} = 5\sqrt{2}\)

tan θ = y/x = 5/5 = 1 

Since the given point lies in the 1st quadrant, 

θ = 45° …[∵ tan 45° = 1] 

∴ the required polar co-ordinates are ( \(5\sqrt{2}\), 45°).

(x, y) = (-1, -1) 

∴ r = \(\sqrt {x^2+y^2} = \sqrt{1+1}=\sqrt2\)

tan θ = \(\frac{y}{x} = \frac {-1}{-1}=1\)

∴ tan θ = tan π/4

Since the given point lies in the 3rd quadrant, 

tan θ = tan (π + π/4) …[∵ tan (n + x) = tan x] 

∴ tan θ = tan (5π/4)

∴ θ = (5π/4) = 225° 

∴ the required polar co-ordinates are ( \(\sqrt2\), 225°).

i. (r, θ) = (3, 90°) 

Using x = r cos θ and y = r sin θ, where (x, y) are the required cartesian co-ordinates, we get 

x = 3cos 90° and y = 3sin 90° 

∴ x = 3(0) = 0 and y = 3(1) = 3 

∴ the required cartesian co-ordinates are (0, 3). 

ii. (r, θ) = (1, 180°) 

Using x = r cos θ and y = r sin θ, where (x, y) are the required cartesian co-ordinates, we get

x = 1(cos 180°) and y = 1(sin 180°) 

∴ x = -1 and y = 0 

∴ the required cartesian co-ordinates are (-1, 0).

(x, y) = (- √ 3, 1) 

∴ r = \(\sqrt{x^2+y^2} = \sqrt{3+1}=\sqrt4=2\)

tan θ = \(\frac{y}{x} = \frac {1}{-\sqrt3}=-tan \frac{\pi}{6}\)

Since the given point lies in the 2nd quadrant, 

tan θ = tan (π-π/6) …[∵ tan (π – x) = – tan x] 

∴ tan θ = tan (5π/6)

∴ θ = (5π/6)= 150° 

∴ the required polar co-ordinates are (2, 150°)

(x, y) = ( 1,√3)

∴ r =\(\sqrt{x^2+y^2} = \sqrt{1+3} = \sqrt4 = 2\)

tan θ = \(\frac{y}{x} = \frac {\sqrt3}{1} =\sqrt3\)

Since the given point lies in the 1st quadrant, 

θ = 60° …[∵ tan 60° = √3] 

∴ the required polar co-ordinates are (2, 60°).

Given : sin (A+B) =1

⇒ Sin(A+B) = sin (90^o) [∵ sin (90^o)=1]

On equating both the sides, we get

A + B = 90^o …(1)

And cos(A - B) = √3/2

⇒ cos(A – B) = cos (30^o) [cos(30^o) = √3/2]

On equating both the sides, we get

A – B = 30^o …(2)

On Adding Eq. (1) and (2), we get

2A = 120^o

⇒ A = 60^o

Now, Putting the value of A in Eq.(1), we get

60^o + B = 90^o

⇒ B = 30^o

Hence, A = 60^o and B = 30^o

Given : sin (A+B) =1

⇒ Sin(A+B) = sin (90^o) [∵ sin (90^o) =1]

On equating both the sides, we get

A + B = 90^o …(1)

And cos (A – B) = 1

⇒ cos(A – B) = cos (0^o) [∵ cos(0^o) = 1]

On equating both the sides, we get

A – B = 0^o …(2)

On Adding Eq. (1) and (2), we get

2A = 90^o

⇒ A = 45^o

Now, Putting the value of A in Eq.(1), we get

45^o + B = 90^o

⇒ B = 45^o

Hence, A = 45^o and B = 45^o

Given : sin(A + B) = √3/2

⇒ Sin(A+B) = sin (60^o) [sin(60^o) = √3/2]

On equating both the sides, we get

A + B = 60^o …(1)

And cos(A - B) = √3/2

⇒ cos(A – B) = cos (30^o) [cos(30^o) = √3/2]

On equating both the sides, we get

A – B = 30^o …(2)

On Adding Eq. (1) and (2), we get

2A = 90^o

⇒ A = 45^o

Now, Putting the value of A in Eq.(1), we get

45^o + B = 60^o

⇒ B = 15^o

Hence, A = 45^o and B = 15^o

Given : sin(A - B) = 1/2

⇒ Sin(A-B) = sin (30^o) [sin (30^o) = 1/2]

On equating both the sides, we get

A - B = 30^o …(1)

And cso (A + B) = 1/2

⇒ cos(A + B) = cos (60^o) [cos (60^o) = 1/2]

On equating both the sides, we get

A + B = 60^o …(2)

On Adding Eq. (1) and (2), we get

2A = 90^o

⇒ A = 45^o

Now, Putting the value of A in Eq.(2), we get

45^o + B =60^o

⇒ B = 15^o

Hence, A = 45^o and B = 15^o

sin (A + B + C) = 1 ⇒ sin (A + B + C) = sin 90° ⇒ A + B + C = 90°            …(i) 

tan (A – B) = \(\frac{1}{\sqrt3}\) ⇒ tan (A – B) = tan 30° ⇒ A – B = 30°                   …(ii) 

sec (A + C) = 2 ⇒ sec (A + C) = sec 60° ⇒ A + C = 60°               …(iii) 

Eq (i) – Eq (iii) ⇒ B = 30° 

∴ From eqn (ii), A – 30° = 30° ⇒ A = 60° 

∴ From eqn (i), 60° + 30° + C = 90° ⇒ C = 0° ⇒ A = 60°, B = 30°, C = 0°.

Given A + B = 90°

∴ \(\sqrt{\text{sin A sec B - sin A cos B}}\)

= \(\sqrt{\text{sin A sec (90°- A) - sin A cos (90° - A)}}\)

= \(\sqrt{\text{sin A cosec A - sin A. sin A}}\) 

= \(\sqrt{1-sin^2 A}\) = \(\sqrt{cos^2 A}\) = cos A.

(b) q

\(\sqrt{p^2-q^2}\) tan \(x\) = \(\sqrt{p^2-q^2\,sin^2\,x.}\,tan\,x\)

= \(\sqrt{p^2(1-sin^2\,x).}\) tan \(x\) = \(p\sqrt{1-sin^2\,x}.\,tan\,x\)   (∵cos²θ + sin²θ = 1)

= \(p\sqrt{cos^2\,x.}\) tan \(x\) = p cos \(x\). tan \(x\) = p sin \(x\) = q

(a) \(\frac7{23}\)

8 tan A = 15, ⇒ tan A = \(\frac{15}{8}\)

Now, \(\frac{sin\,A-cos\,A}{sin\,A+cos\,A}\) = \(\frac{\frac{sin\,A}{cos\,A}-\frac{cos\,A}{cos\,A}}{\frac{Sin\,A}{cos\,A}+\frac{cos\,A}{cos\,A}}\)

= \(\frac{tan\,A-1}{tan\,A+1}\)

Now, substitute the value of tan A.

(c) \(\sqrt{\frac{a^2+b^2}{b}}\)

Given, \(\frac{sin\,\theta}{cos\,\theta}=\frac{a}{b}\) ⇒ tan θ = \(\frac{a}{b}\)

Also, sec² θ = 1 + tan²θ ⇒ 2 sec θ = \(\sqrt{1+tan^2\,\theta}\)

⇒ sec θ = \(\sqrt{1+\frac{a^2}{b^2}}=\) \(\sqrt{\frac{b^2+a^2}{b^2}}=\)\(\sqrt{\frac{a^2+b^2}{b}}\)