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Given:

Our given expression is tan(2x + y) tan(2x – y) = 1

Formula used:

When A + B = 90° then, tanA tanB = 1 and vice versa

Calculation:

Our given expression is tan(2x + y) tan(2x – y) = 1

⇒ 2x + y + 2x – y = 90°

⇒ 4x = 90°

⇒ 2x = 45°

Now, tan2x = tan45° = 1

∴ The value of tan45° is 1

Given:

Our given expression is 7 – 2sin²θ – 11cosθ = 0

Formula used:

cos²θ + sin²θ = 1

Calculation:

Our given expression is 7 – 2sin²θ – 11cosθ = 0

⇒ 7 – 2(1 – cos²θ) – 11cosθ = 0

⇒ 7 – 2 + 2cos²θ – 11cosθ = 0

Put cosθ = t, then our equation becomes

⇒ 2t² – 11t + 5 = 0

⇒ (t – 5)(t – 1/2) = 0

⇒ t = 5 or t = ½

Here t = 5 is not possible, so t = ½

⇒ cosθ = ½

∴ The value of θ is 60° 

Concept:

In triangle ABC, A + B + C =180° 

If a, b,c are in AP then 2b = a + c

tan 60 = √3

Calculation:

Here, A, B, C are in AP, so 2B = A + C = 30 + C         .....(1)

ABC is a triangle, so A + B + C =180° ⇒ B = 150 - C .....(2)

Add (1) and (2), we get 

3B = 180 ⇒ B = 60° 

So, C = 90° 

2sinA + 3tanB - 4cosC = 2(sin 30) + 3 tan (60) - 4 cos 90

= 2(1/2) + 3 √3

= 1 + 3√3

Hence, option (4) is correct. 

\(2{\tan ^{ - 1}}\left( {\frac{2}{3}} \right) + {\cos ^{ - 1}}\left( {\frac{{12}}{{13}}} \right)\)

\( = {\tan ^{ - 1}}\left( {\frac{2}{3}} \right) + {\tan ^{ - 1}}\left( {\frac{2}{3}} \right) + {\tan ^{ - 1}}\left( {\frac{5}{{12}}} \right)\;\)

\(= {\tan ^{ - 1}}\left[ {\frac{{\frac{2}{3} + \frac{2}{3}}}{{1 - \frac{2}{3} \times \frac{2}{3}}}} \right] + {\tan ^{ - 1}}\left( {\frac{5}{{12}}} \right)\)

\( = {\tan ^{ - 1}}\left[ {\frac{{\frac{4}{3}}}{{1 - \frac{4}{9}}}} \right] + {\tan ^{ - 1}}\left( {\frac{5}{{12}}} \right)\;\)

\( = {\tan ^{ - 1}}\left[ {\frac{{\frac{4}{3}}}{{\frac{5}{9}}}} \right] + {\tan ^{ - 1}}\left( {\frac{5}{{12}}} \right)\)

\( = {\tan ^{ - 1}}\left[ {\frac{{36}}{{15}}} \right] + {\tan ^{ - 1}}\left[ {\frac{5}{{12}}} \right]\)

\( = {\tan ^{ - 1}}\left[ {\frac{{12}}{5}} \right] + {\tan ^{ - 1}}\left[ {\frac{5}{{12}}} \right]\)

\( = {\tan ^{ - 1}}\left[ {\frac{{12}}{5}} \right] + {\cot ^{ - 1}}\left[ {\frac{{12}}{5}} \right]\)

\( = \frac{\pi }{2}\left[ {{{\tan }^{ - 1}}x + {{\cot }^{ - 1}}x = \frac{\pi }{2}} \right]\)

Given :

(sec θ + tan θ)/(sec θ - tan θ) = 209/79

Calculations :

\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaaeaaaaaaaaa8 % qadaWcaaWdaeaapeGaci4CaiaacwgacaGGJbGaeqiUdeNaey4kaSIa % ciiDaiaacggacaGGUbGaeqiUdehapaqaa8qacaWGZbGaamyzaiaado % gacqaH4oqCcqGHsislcaWG0bGaamyyaiaad6gacqaH4oqCaaaaaa!4A3F! \frac{{\sec θ + \tan θ }}{{secθ - tanθ }}\)sec θ + tan θ)/(sec θ - tan θ) = 209/79

Cross multiply both the sides

79 sec θ + 79 tan θ = 209 sec θ - 209 tan θ 

⇒ 288 tan θ = 130 sec θ 

⇒ sin θ/cos θ = (130/288) (1/cos θ)

⇒ sin θ = 65/144

∴The value of sinθ will be 65/144

Given:

sec θ + tan θ = 4      ----(i)

Formula used:

sec² θ – tan² θ = 1

(a² – b²) = (a + b)(a – b)

sec θ = Hypotenuse/Base

sin θ = Perpendicular/Hypotenuse

Calculations:

Using the given identities,

⇒ (sec θ + tan θ)(sec θ – tan θ) = 1

⇒ 4 × (sec θ – tan θ) = 1

⇒ (sec θ – tan θ) = 1/4      (ii)

Adding (i) and (ii),

⇒ 2sec θ = 17/4

⇒ sec θ = 17/8

sin θ = √(17² – 8²)/17

⇒ sin θ = 15/17

∴ The value of sin θ is 15/17

Explanation:

θ = sin-1 (sin (-600°))

θ = sin-1 (sin (-600°))

θ = sin-1 (sin (-600°)) ----------(1)

Since, sin (θ + (n × 360°)) = sin θ

where n is a natural number

Continuing from equation (1)

θ = sin-1 (sin (-600°))

θ = sin-1 (sin (-600° + (2 × 360°)))

θ = sin-1 (sin (-600° + 720°))

θ = sin-1 (sin 120°)

θ = \(sin^{-1}(\frac{\sqrt{3}}{2})\)

θ = π/3 

Note: Range of inverse sine function is [-π/2 to π/2].

GIVEN:

sec A + tan A = m

sec A. tan A = n

sec⁴ A – tan⁴ A = m²/2

CONCEPT:

Trigonometry

CALCULATION:

sec A + tan A = m

⇒ (sec A + tan A)² = m²

⇒ sec² A + tan² A + 2 sec A. tan A = m²

⇒ sec² A + tan² A + 2n = m²

⇒ sec² A + tan² A = m² – 2n

Now,

sec⁴ A – tan⁴ A = m²/2

⇒ (sec² A – tan² A) (sec² A + tan² A) = m²/2   [sec² A – tan² A = 1]

⇒ 1 (sec² A + tan² A) = m²/2

⇒ m² – 2n = m²/2

⇒ 2m² – 4n = m²

⇒ m² = 4n

Concepts:

If \(\rm \frac A B = \frac C D\), property of componendo and dividendo is given by, \(\rm \frac{A+B}{A-B}=\frac{C+D}{C-D}\)

Formula:

• sin (x + y) = sin x cos y + cos x sin y
• sin (x – y) = sin x cos y - cos x sin y

Calculation:

Given:

\(\rm \frac{{\sin \left( {x\; + \;y} \right)}}{{\sin \left( {x\; - \;y} \right)}} = \frac 53\)

\(\Rightarrow\rm \;\frac{{\sin x\cos y\; + \;\cos x\sin y}}{{\sin x\cos y\; - \;\cos x\sin y}} = \frac 53\)

Using componendo and dividendo formula, we get

\(\rm \Rightarrow \frac{{(\sin x\cos y + \cos x\sin y) + (\sin x\cos y - \cos x\sin y)}}{{(\sin x\cos y + \cos x\sin y) - (\sin x\cos y - \cos x\sin y)}} = \frac{5+3}{5-3}\)

\(\rm \Rightarrow \;\frac{{2{\rm{\;sin}}x\cos y}}{{2\cos x\sin y}} = \frac{8}{2}\)

(b) In ΔABC, angles A,B,C are in A.P., 

B – A = C – B 

2B = A + C      ...(i) 

also, A + B + C = 180° (Angle sum Property) 

2 B + B = 180° 

B = 60° 

Hence, sin B = sin60° =  √3/2

Given:

x = a (sin 0 + cos 0)

y = b (sin 0 - cos 0)

Formula used:

sin 0 = 0

cos 0 = 1

Calculation:

x = a(0 + 1) = a

y = b(0 - 1) = - b

\(\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}\) = \(\dfrac{a^2}{a^2}+\dfrac{(- b)^2}{b^2}\)

⇒ \(\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}\) = 1 + 1

∴ \(\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}\) = 2

Formula used:

The maximum value of a sin x + b cos x = √(a² + b²)

Calculation:

The maximum value of 2 sin 0 + 3 cos 0 = √(2² + 3²) = √13

∴ Maximum value of 2 sin 0 + 3 cos 0 = √13

Given:

4 – 2sin2 θ – 5cos θ = 0

Identity used:

sin²θ = 1 – cos²θ 

Calculation:

4 – 2sin2 θ – 5cos θ = 0

⇒ 4 – 2 × (1 – cos2θ) – 5cosθ = 0

⇒ 4 – 2 + 2cos2θ – 5cosθ = 0

⇒ 2cos2θ – 5cosθ + 2 = 0

⇒ 2cos2θ – 4cosθ – cosθ + 2 = 0

⇒ 2cosθ × (cosθ – 2) – 1 × (cosθ – 2) = 0

⇒ (2cosθ – 1) × (cosθ – 2) = 0

⇒ cosθ = 1/2 and cosθ = 2

Rejecting cosθ = 2 as 0° < θ < 90°

So, cosθ will be 1/2 

⇒ θ = 60° 

The value of cos θ + tan θ = cos60° + tan60° 

⇒ (1/2) + √3

∴ The value of cos θ + tan θ is \(\dfrac{1+2\sqrt{3}}{2}\)

Given:

(sin θ + cos θ)2 = 2

Formula used:

(A + B)² = A² + B² + 2AB

sin2θ + cos2θ = 1

sin2θ = 2sinθcosθ

sin90º = 1

Calculation:

(sin θ + cos θ)2 = 2

⇒sin²θ + cos² +  2sinθcosθ = 2

⇒1 + 2sinθcosθ = 2

⇒2sinθcosθ = 2 - 1

⇒2sinθcosθ = 1

⇒sin2θ = 1

⇒sin2θ = sin90º

⇒2θ = 90º

⇒θ = 45º

⇒θ = π/4

Given:

The given condition is √3sinθ = 3 cosθ

Formula Used:

Basic concept of trigonometric ratio and identities

We know that

tanθ = sinθ/cosθ

tan60° = √3 and Sec60° = 2

Calculation:

By rearranging the expression

∴ √3sinθ = 3 cosθ

⇒ Sinθ/cosθ = 3/√3

⇒ tanθ = 3/√3

Now, by rationalizing the denominator

∴ tanθ = √3

Also, tan60° = √3

∴ tanθ = tan60°

⇒ θ = 60°

Now, the value of sec²θ – 2

∴ sec²(60°) – 2

⇒ 4 – 2 = 2

Hence, option (2) is correct

Given:

It is given that sin3θ.sec6θ = 1

Formula Used:

Basic concept of trigonometric ratio and identities

We know that

Secθ = 1/cosθ

Cos(90 - θ) = sin θ

tan30° = 1/√3

Calculation:

As we know that reciprocal of secθ is equal to cosθ

∴ Sin3θ/cos6θ = 1

⇒ sin3θ = cos6θ

∵ Cos(90 - θ) = sin θ

∴ sin3θ = sin(90 - 6θ)

⇒ 3θ = 90 - 6θ

⇒ θ = 10°

Now, 3tan²3θ

∴ 3tan²3θ = 3tan²30° = 3 × (1/√3)² = 1

Hence, option (1) is correct

Given:

\(\frac{{\sin x + \cos x}}{{\sin x - \cos x}} = \frac{6}{5}\)

Concept Used:

Componendo & dividendo

a/b = c/d

then (a + b)/(a - b) =(c + d)/(c - d)

Calculation:

\(\frac{{\sin x + \cos x}}{{\sin x - \cos x}} = \frac{6}{5}\)

By using componendo & dividendo

(sinx + cosx + sinx - cos x)/(sinx + cosx - sinx + cosx)  = (6 + 5)/(6 - 5)

(2sinx/2cosx) = (sinx/cosx) = 11/1

tan²x = (sin²x/cos²x) = 11²/1² = 121

tan²x + 1 = 121 +1 =122

tan²x - 1 = 121 - 1 = 120 

\(\frac{{{{\tan }^2}x + 1}}{{{{\tan }^2}x - 1}}\) = 122/120 = 61/60

∴ The value of \(\frac{{{{\tan }^2}x + 1}}{{{{\tan }^2}x - 1}}\) is 61/60

Given:

It is given that sec (A + B) = √2 and sin (A - B) = 1

Formula Used:

Basic concept of trigonometric ratio and identities

We know that

Sec 45° = √2 and sin 90° = 1

Calculation:

The equation sec (A + B) = √2 can be written as

∵ Sec 45° = √2

∴ Sec (A + B) = Sec 45°

⇒ A + B = 45°    ---(1)

And sin (A - B) = 1 so it can be written as

∵ Sin 90° = 1

∴ sin (A - B) = Sin 90°

⇒ A - B = 90°     ---(2)

By equation (1) and (2) we get

∴ A = 67.5° and B = -22.5 °

Now, we have to find the value of 2A + B

∴ 2 × 67.5 + (-22.5) = 112.5°

Hence, option (1) is correct

Concept:

• sin² θ + cos² θ = 1.
• sin 2θ = 2 sin θ cos θ.

Calculation:

Consider the expression \(\rm \cot{\frac{A}{2}}+ \tan {\frac{A}{2}}\).

= \(\rm \frac{\cos\frac{A}{2}}{\sin\frac A2}+ \frac{\sin \frac{A}{2}}{\cos\frac{A}{2}}\)

= \(\rm \frac{\cos^2\frac{A}{2}+\sin^2\frac{A}{2}}{\sin\frac A2\cos\frac{A}{2}}\)

= \(\rm \frac{1}{\sin\frac A2\cos\frac{A}{2}}\)

= \(\rm \frac{2}{2\sin\frac A2\cos\frac{A}{2}}\)

= \(\rm \frac {2}{\sin A}\)

= 2 cosec A

Given:

The given expression is sec279° – (1/tan211°)

Formula Used:

Basic concept of trigonometric ratio and identities

We know that

Sec² θ – tan² θ = 1

1/tan θ = cot θ

Cot (90 – θ) = tan θ

Calculation:

The given expression sec²79° – (1/tan²11°) can be written as sec²79° –  cot²11°

∵ 1/tan θ = cot θ

Now, sec²79° – cot²11° = sec²79° – cot²(90 – 79°) = sec²79° – tan² 79°

∵ sec²θ – tan²θ = 1

∴ sec²79° – tan²79° = 1

Hence, option (2) is correct

Given:

It is given that sec α × sin 59 ° = 1

Formula Used:

Basic concept of trigonometric ratio and identities

We know that

1/cosec θ = sin θ

Cosec (90 - θ) = sec θ

Calculation:

We know that the reciprocal of sin θ is cosec θ

∴ Sec α × sin 59 ° = 1 can be written as sec α × 1/cosec 59 ° = 1

⇒ Sec α = cosec 59 °

⇒ Sec α = cosec (90 - 31) °

∵ Cosec (90 - θ) = sec θ

∴ Sec α = sec 31°

⇒ α = 31°

Hence, option (1) is correct

Concept:

• sin 2x = 2 sin x cos x
• cos 2x = cos² x - sin² x

Calculation:

Consider, tan A + sec A.

= \(\rm \frac {\sin A}{\cos A} + \frac {1}{\cos A}\)

= \(\rm \frac {\sin A+1}{\cos A}\)

= \(\rm \frac {2\sin \frac{A}{2}\cos \frac{A}{2}+\sin^2\frac{A}{2}+\cos^2\frac{A}{2}}{\cos^2 \frac{A}{2}-\sin^2\frac{A}{2}}\)

= \(\rm \frac {\left(\sin \frac{A}{2}+\cos \frac{A}{2}\right)^2}{\left(\sin \frac{A}{2}+\cos \frac{A}{2}\right)\left(\cos \frac{A}{2}-\sin \frac{A}{2}\right)}\)

= \(\rm \frac {\sin \frac{A}{2}+\cos \frac{A}{2}}{\cos \frac{A}{2}-\sin \frac{A}{2}}\)

Dividing by \(\rm \sin\frac{A}{2}\), we get:

= \(\rm \frac {\cot\frac{A}{2}+1}{\cot\frac{A}{2}-1}\)

Using \(\rm \cot\frac{\pi}{4}=1\), it can be written as;

= \(\rm \frac {\cot\frac{\pi}{4}\cot\frac{A}{2}+1}{\cot\frac{A}{2}-\cot\frac{\pi}{4}}\)

= \(\rm \cot \left( \frac {\pi} 4 - \frac{A}{2} \right)\)

Given:

 tan2x° = cot(x+6)°

Concept used:

tan(90° - θ) = cotθ 

Calculation:

 tan2x° = cot(x+6)°

⇒  tan2x° = tan[90° - (x+6)°]                           [tan(90° - θ) = cotθ]

⇒ 2x° = 90° - (x+6)°

⇒ 2x° = 90° - x° - 6° 

⇒ 3x° = 84° 

⇒ x = 28° 

∴ The value of x is 28°.

Given:

sin x + cos x = √3 cos x

Concept used:

Rationalization method used.

Formula used:

Tanθ = Sinθ/Cosθ

 a² – b² = (a + b) × (a  – b)

Calculation:

sin x + cos x = √3 cos x

⇒ (sin x + cos x)/cos x = √3

⇒ (sin x/cos x + cos x/cos x ) = √3

⇒ tan x + 1 = √3

⇒ tan x = √3 – 1

⇒ 1/cot x = √3 – 1

⇒ cot x = 1/(√3 – 1)

⇒ cot x = (√3 + 1)/[(√3 – 1) × (√3 + 1)]

⇒ cot x =  (√3 + 1)/[(√3)² – 1]

⇒ cot x =  (√3 + 1)/(3  – 1)

⇒ cot x =  (√3 + 1)/2

∴ The value of cot x is (√3 + 1)/2.

Given:

sin35° × cos55°

Concept used:

sin(90° - θ) = cosθ 

cos(90° - θ) = sinθ 

Calculation:

sin35° × cos55°

⇒ sin35° × cos(90° - 35°)

⇒ sin35° × sin35°

⇒ sin²35° 

∴ The value is sin235°.

Given:

3b cosecθ = a secθ 

3a secθ - b cosecθ = 8

Formula used:

sin²θ + cos²θ = 1

Calculation:

∵ 3b cosecθ = a secθ

⇒ 9b cosecθ = 3a secθ      ------(1)

∵ 3a secθ - b cosecθ = 8

⇒ 9b cosecθ - b cosecθ = 8      ------(From 1)

⇒ 8b cosecθ = 8

⇒ b cosecθ = 1

⇒ b = 1/cosecθ      ------(2) 

⇒ b = sinθ      ------(3)

∵ 3a secθ - b cosecθ = 8

⇒ 3a secθ - (cosecθ/cosecθ) = 8

⇒ 3a secθ - 1 = 8

⇒ 3a secθ = 9

⇒ a secθ = 3

⇒ a = 3/secθ

⇒ a = 3cosθ      ------(4)

Putting the values of 'a' and 'b' from (3) and (4) in 9b² + a² 

= 9 (sinθ)² + (3cosθ)²

= 9 sin²θ + 9 cos²θ 

= 9(sin²θ + cos²θ)

= 9 × 1

= 9

(a) sinA + sin²A = 1 

sinA = cos²A 

sin²A = cos⁴A 

cos⁴A = 1– cos²A 

cos²A + cos⁴A = 1