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x = 5 + 6cosec θ and y = 3 + 8cot θ 

∴ x – 5 = 6cosec θ and y – 3 = 8cot θ

∴ cosec θ = \(\frac{x-5}{6}\) and cot θ \(\frac{y-3}{8}\)

We know that,

cosec² θ – cot² θ = 1

 ∴ \((\frac{x-5}{6})^2 - (\frac{y-3}{8})^2=1\)

x = 6cosec θ and y = 8cot θ 

∴ cosec θ = and cot θ = 

We know that, 

cosec² θ – cot² θ =

∴ \((\frac{x}{6})^2 - (\frac{y}{8})^2=1\)

∴ \(\frac{x^2}{36} - \frac{y^2}{64}=1\)

16x² – 9y² = 576

x = 3sec θ, y = 4tan θ

∴ sec² θ = x/3 and tan θ = y/4

We know that,

sec²θ - tan²θ = 1

∴ \((\frac{x}{3})^2 - (\frac{y}{4})^2 = 1\)

∴ \(\frac {x^2}{9} - \frac{y^2}{16}=1\)

∴ 16x² – 9y² = 144

Given, \(\frac{sin A}{3} = \frac{sin B}{4} = \frac{1}{5}\)

∴ sin A = \(\frac{3}{5}\) and sin B = \(\frac{4}{5}\)

We know that,

cos² A = 1 - sin² = 1 - \((\frac{3}{5})^2 = 1 - \frac{9}{25} = \frac{16}{25}\)

∴ Cos A = ± 4/5

Since A lies in the second quadrant, 

cos A < 0 

∴ Cos A = – 4/5 

Sin B = 4/5 

We know that,

cos² B = 1 - sin² B = 1 - \((\frac{4}{5})^2 = 1 - \frac{16}{25} = \frac {9}{25}\)

∴ Cos B = ± 4/5

Since B lies in the second quadrant, cos B < 0 

∴ cos B = - 3/5

∴ 4cos A + 3 cos B = \(4(-\frac{4}{5}) + 3 (-\frac{3}{5})\) 

= \(-\frac{16}{5} - \frac {9}{5} = -\frac{25}{5}= -5\)

L.H.S. = sin2A + sin2B + sin2C 

= 2sin(A + B) · cos(A – B) + 2sinCcosC 

= 2sinC.cos(A – B) + 2 sinC cosC 

= 2sinC [cos(A – B) – cos(A + B)] 

= 2 sinC[-2 sinA · sin(-B)] 

= 4sinA . sinB . sinC 

∵ sin(-B) = R.H.S. = SinB.

Given A + B + C = \(\frac{\pi}{2}\); A + B = 90° – C; tan(A + B) = tan(90° – C)

\(\frac{tanA + tanB}{1 - tanAtanB} = cotc \)

\(\frac{1}{tanC}\)

⇒ tanA tanC + tanB tanC = 1 – tanA · tanB 

⇒ tan A tanB + tanB tanC + tanC tanA = 1.

Given A + B + C = π 

⇒ A + B = π – C 

tan(A + B) = tan(180 – C) 

\(\frac{tanA + tanB}{1 - tanA.tanB} = - \frac{tanc}{1}\)

tanA + tanB = – tanc – tanB tanB · tanc 

tanA + tanB + tanC = tanA tanB tanC

L.H.S. = cos2A .\(\frac{1}{secA}\) + sin 2A .\(\frac{1}{cosA}\) 

= cos2A . cosA + sin2A . sinA 

= cos(2A – A) = cosA = R.H.S.

Correct option is: B) \(\frac 18\)

We have Tan \(\theta\) + Sec  \(\theta\)  = 8

\(\therefore\) ( sec \(\theta\) + tan \(\theta\)) (sec \(\theta\) - tan \(\theta\)) = 8 (sec \(\theta\) - tan \(\theta\))

(on multiplying both sides by sec \(\theta\) - tan \(\theta\))

= \(sec^2\theta-tan^2 \theta\) = 8 (sec \(\theta\) - tan \(\theta\)) (\(\because\) (a+b) (a-b) = \(a^2 - b^2\))

= 8 (sec \(\theta\) - tan \(\theta\)) = 1 (\(\because\) \(sec^2\theta-tan^2 \theta\) = 1)

= sec \(\theta\) - tan \(\theta\) = \(\frac 18\) 

Correct option is: B) \(\frac{1}{8}\)

Given, 

√3 sin θ = cos θ 

sin θ / cos θ = 1/√3

tan θ = 1/ √3 …(the value of tan 30° is 1/√3)

Therefore, tan θ = tan 30°θ = 30°

Given, 

cos B = 3/5 and 

A + B = 90° 

So we have, 

B = 90° - A 

So, 

cos B = cos (90° - A) = 3/5 

∵ cos B = sin A 

Therefore, 

sin A = 3/5

Sin 3x =√3/2 Sin 3x = sin 60°

Equating angles we get,

3x = 60°
x = 20°

Given: sin (A + B) = 1 and cos (A – B) = 1 

sin(A+B) = 1 

Also, we know, sin 90° = 1 

⇒ sin(A+B) = sin 90° 

or (A+B) = 90° .............(1) 

Now, cos (A – B) = 1 

And, we know, cos 0° = 1 

⇒ (A - B) = 0° .............(2) 

On solving both equations (1) and (2) , 

we get 

2A = 90° 

or A = 90°/2 

or A = 45°

Similarly, B = 45°

L.H.S. = cosA + cos A(120° – A) + cos(120° + A) 

= cosA + 2cos 120°. cosA 

= cosA + 2cosA (-cos60°) {∵ cos 120° = cos (180° - 60°) = -cos 60°}

= cos A – 2 cosA . \(\frac{1}{2}\) = 0 = R.H.S.

(a) x – y = 0

tan² y cosec² x – 1 = tan² y 

⇒ tan² y cosec² x – tan² y = 1 

⇒ tan² y (cosec² x – 1) = 1 

⇒ tan² y . cot² x = 1 ⇒ cot² x = cot² y ⇒ x = y ⇒ x – y = 0.

Given : sin 60° cos 30° + cos 60°sin 30°

To find : The value of sin 60° cos 30° + cos 60°sin 30°

Solution : Use the values:

sin 30 = \(\frac{1}{2}\), sin 60 = \(\frac{\sqrt3}{2}\),cos30 = \(\frac{\sqrt3}{2}\) and cos 60 = \(\frac{1}{2}\) 

sin 60° cos 30° + cos 60°sin 30°

Solve, \(\frac{\sqrt3}{2}\) x \(\frac{\sqrt3}{2}\) + \(\frac{1}{2}\) x \(\frac{1}{2}\)

= \(\frac{3}{4}\) + \(\frac{1}{4}\) = 1 

Hence the value of sin 60° cos 30° + cos 60°sin 30° is 1.

tan A = \(\frac{15}8\)

⇒ sin A = \(\frac{15}{17}\) and cos A = \(\frac{8}{17}\)

∴ sin A - cos A = \(\frac{15}{17}\) - \(\frac{8}{17}\) = \(\frac{7}{17}\)