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Calculation:

{[Tan A/(1 – Cot A)]+ [Tan A/(1 – Cot A)]} × (1 – Cot 2A)

⇒ {[Tan A/(1 – Cot A) ]+ [Tan A/(1 – Cot A)]} × (1 – CotA) × (1 + CotA)

⇒ TanA × (1 + CotA) + TanA × (1 – CotA)

⇒ TanA × (1 + CotA + 1 – CotA)

⇒ TanA × 2

∴ 2TanA

Given

4 cos2 θ - 3 sin2 θ + 2 = 0

Formula:

sin2θ + cos2θ = 1

tan2θ = sin2θ/cos2θ

Calculation:

4 cos2θ - 3 sin2θ + 2 = 0

⇒ 4 cos²θ - 3 (1 - cos²θ) + 2 = 0

⇒ 4 cos²θ - 3 + 3 cos²θ + 2 = 0

⇒ 7 cos²θ - 1 = 0

⇒ 7 cos²θ = 1

⇒ cos²θ = 1/7

sin²θ + cos²θ = 1

⇒ sin²θ = 1 - 1/7

⇒ sin²θ = 6/7

Now,

tan²θ = sin²θ/cos²θ

⇒ tan²θ = (6/7)/(1/7)

⇒ tan²θ = 6

∴ tanθ = √6

Given 

cos 57° + sin 27° 

Formula Used 

cos(90° - θ) = sinθ 

sinC + sinD = 2 × sin(C + D)/2 × cos(C - D)/2

Calculation 

cos(90° - θ) = sinθ 

⇒ cos 57° + sin 27° = cos (90° - 33°) + sin 27° 

⇒ sin 33° + sin 27° 

⇒ 2sin (33° + 27°)/2 × cos(33° - 27°)/2

⇒ 2sin30° × cos 3° 

⇒ 2 × 1/2 cos 3° 

⇒ cos 3°

∴ The required answer is cos3°

Given:

cot 35°cot 40°cot 45°cot 50°cot 55°

Concept used:

cot 45^° = 1

tan θ × cot θ = 1

tan A = cot B, if A + B = 90^° 

Calculation:

cot 35°cot 40°cot 45°cot 50°cot 55°

⇒ cot 35^° cot 40^° cot 45^° tan 40^° tan 35^° 

⇒ cot 35^° tan 35^° cot 40^° tan 40^° cot 45^° 

⇒ cot 45^° = 1

∴ The correct answer is 1.

We know that  
sin (x + y) = sin x cos y + cos x sin y ... (1)  
Now cos²x = 1 – sin²x = 1 – 9/25 = 16/25  
Therefore cos x = ± 4/5.  
Since x lies in second quadrant, cos x is negative.  
Hence cos x = −4/5  
Now sin²y = 1 – cos²y = 1 – 144/169 = 25/169  
i.e. sin y = ± 5/13.  
Since y lies in second quadrant, hence sin y is positive. 

Therefore, sin y =5/13. 

Substituting the values of sin x, sin y, cos x and cos y in (1), we get  
sin(x + y) 3/5 × (-12/13) + (−4/5) × 5/13 = (-36/65) –(20/65) = -56/65

Therefore, sin(x+y) = -56/65

Answer:

tan(19π/3)=√3=1.732

Explanation:

Every trigonometric function has a cycle of 2π, i.e. after 2π, the function is repeated. In fact functions tan and cot repeat after every π.

As tan(19π/3)=tan[(18π+π)/3]=tan(6π+π/3)=tan(π/3)

(as 6π is a multiple of π which is cycle for function tan)

and as tan(π/3)=√3, we have

tan(19π/3)=√3=1.732

Given:

sinA= 7/25 

Concept used:

sinA = P/H 

tanA = P/B 

cotA = B/P 

cosA = B/H 

Phythagoras theorem 

H² = P² + B² 

Where, 

P → Perpendicular 

B → Base 

H → Hypotenus 

Calculations:

sinA = 7/25 = P/H 

P = 7 and H = 25 

According to phythagoras theorem 

⇒ 25² = 7² + B² 

⇒ B² = 625 – 49 

⇒ B = √576 = 24 

tanA = P/B = 7/24 

cotA = 24/7 

cosA = 24/25 

48 tan A - 21 cot A + 50 cos A.

⇒ 48 × (7/24) – 21 × (24/7) + 50 × (24/25) 

⇒ 14 – 72 + 48 

⇒ -10

∴ The correct answer is -10 

Given

ysin30° = xsin45°

Formula

sin45° = 1/√2

sin30° = 1/2

Calculation

y × (1/2) = x × (1/√2)

⇒ y/x = (2/√2) = √2

⇒ y4/x4 = (√2)4

Given - 

(tan9° - tan27° - tan63° + tan81°)

Concept used - 

If tanθ + cotθ = x

then, x = (2/sin2θ) = 2 cosec2θ 

Formula used - 

sin(90° - θ) = cosθ 

tan(90° - θ) = cotθ 

cot(90° - θ) = tanθ 

tanθ = (sinθ/cosθ)

cotθ = (cosθ/sinθ) 

sinC - sinD = 2 cos{(C + D)/2} sin{(C - D)/2}

Solution - 

(tan9° - tan27° - tan63° + tan81°) 

⇒ tan9° - tan27° - tan (90° - 27°) + tan (90° - 9°)

⇒ tan9° + cot9° - (tan27° + cot27°)

⇒ (2/sin18°) - (2/sin54°)

⇒ 2 × {(sin54° - sin18°)/sin18°sin54°)}

⇒ 2 × 2 (cos36° × sin18°)/(sin18° sin54°)

⇒ 4

∴ Ans = 4.

Given:

Sin (α - β) = 2√2/3 and cosec (α + β) = 2√2/3

Concept Used:

Basic concept of trigonometric ratio and identities

We know that

(α² – β²) = (α - β) × (α + β)

sin ^-1a × cosec ^-1a = 1

Calculation:

It is given that sin (α - β) = 2√2/3

∴ (α - β) = sin ^-1 2√2/3      ---(1)

And cosec (α + β) = 2√2/3

∴ (α + β) = cosec ^-1 2√2/3      ---(2)

By multiplying equation (1) and (2)

∴ (α - β) × (α + β) = sin ^-1 2√2/3 × cosec ^-1 2√2/3

⇒ (α² – β²) = 8/9

Now, tan (α² – β²) = tan 8/9

Hence, option (1) is correct

1) sec ∝ Range is [- ∞, 1] ⋃ [1, ∞]

sec ∝ = 1/5 (Not possible)

2) tanβ range is [-∞, ∞]

tanβ = 15 (possible)

3) cosec γ Range is [-∞, 1] ⋃ [1, ∞]

cosec γ = 1/3 (Not possible)

4) cos δ Range is [-1, 1]

cos δ = 4 (Not possible)

Given:          

2cos²θ – 5cosθ + 2 = 0

Formula used:

Cos60° = 1/2

Cosec60° = 2/(√3)

Cot60° = 1/√3

Calculation:

2cos2θ – 5cosθ + 2 = 0

⇒ 2cos2θ – 4cosθ – cosθ + 2 = 0

⇒ 2cosθ(cosθ – 2) –1(cosθ – 2) = 0

⇒ (cosθ – 2)(2cosθ – 1) = 0

⇒  (cosθ – 2) = 0 or, (2cosθ – 1) = 0

⇒ cosθ = 2 or, cosθ = 1/2

0º < θ < 90º 

So, cosθ = 1/2 = cos60° 

⇒ θ = 60° 

1/(cosecθ + cotθ) = 1/(cosec60° + cot60°)

⇒ 1/[2/(√3) + 1/√3]

⇒ √3/3

∴ The value of 1/(cosecθ + cotθ) is √3/3.

Formula used:

tanθ = cot(90° – θ)

cotθ = 1/tanθ 

Calculation:

tan75° = cot(90° – 75°)

⇒ tan75° = cot15° 

cot15° = 1/tan15° 

We have to find the value of 1 + tan15°tan75° 

⇒ 1 + tan15° × cot15° 

⇒ 1 + tan15°/tan15° 

⇒ 1 + 1 = 2

∴ The value of the given expresison is 2

Given:

Cosec A = 25/7

Formula Used:

Cosec A = Hypotenuse/Perpendicular

Tan A = Perpendicular/Base

Calculation:

Cosec A = Hypotenuse/Perpendicular = 25/7

From here we deduce that Hypotenuse = 25 Units and Perpendicular = 7 Units

Using Pythagoras Theorem in right angle triangle ABC, where AB is perpendicular, BC is Base, and AC is Hypotenuse.

AC² = AB² + BC²

⇒ (25)² = (7)² + (BC)²

⇒ 625 = 49 + BC²

⇒ BC² = 625 -49 = 576

⇒ BC = √576 = 24 Units (Base)

Tan A = 7/24

∴ The value of Tan A is 7/24

Given:

A + B = 90° and SinA = 3/5 

Concept Used:

Sinθ = Perpendicular/Hypotenuse

tanθ = Perpendicular/Base

Hypotenuse² = Perpendicular² + Base²

tan(90° - θ) = cotθ 

Calculation:

SinA = 3/5 = Perpendicular/Hypotenuse

⇒ Perpendicular = 3k and Hypotenuse = 5k  [Where, k is a constant]

Base = √(Hypotenuse² - Perpendicular)

⇒ Base = √(5k)² - (3k)²

⇒ Base = 4k

tanA = Perpendicular/Base

⇒ tanA = 3k/4k = 3/4

⇒ tan(90° - B) = 3/4

⇒ cotB = 3/4

⇒ tanB = 4/3

∴ tanB = 4/3

Given:

√(cosecx – 1)/√(cosecx + 1)  

Trigonometry identities used:

sin²x + cos²x = 1

Formula used:

 a² – b² = (a + b) (a – b)

Calculation:

√(cosecx – 1)/√(cosecx + 1) 

(Here cosecx = 1/sinx

√[(1/sinx) – 1]/√[(1/sinx) + 1]

= √(1 – sinx)/√(1 + sinx)

Rationalization

=  [√(1 – sinx)/√(1 + sinx)] ×  [√(1 – sinx)/√(1 – sinx)]   

= √(1 – sinx)²/√(1 + sinx) (1 – sinx)

Here a² – b² = (a + b) (a – b)

= (1 – sinx)/√(1 – sin²x)

Here 1 – sin²x = cos²x

= (1 – sinx)/√cos²x

= (1 – sinx)/cosx

= (1/cosx) – (sinx/cosx)

Here 1/cosx = secx, sinx/cosx = tanx

= secx – tanx    

Given:

\(\frac{{\cos \,{{29}^ \circ }\cos ec\,{{61}^ \circ }\tan \,{{45}^ \circ } + 2\sin \,{{35}^ \circ }\sec \,{{55}^ \circ }}}{{3{{\sin }^2}\,{{42}^ \circ } + 3{{\sin }^2}\,{{48}^ \circ }}}\)

Trigonometry properties:

cosecθ = 1/sinθ 

sin(90 – θ ) = cosθ 

secθ = 1/cosθ 

cos(90 – θ ) = sinθ 

sin²θ + cos²θ = 1

tan 45^o = 1

Calculation:

\(\frac{{\cos \,{{29}^ \circ }\cos ec\,{{61}^ \circ }\tan \,{{45}^ \circ } + 2\sin \,{{35}^ \circ }\sec \,{{55}^ \circ }}}{{3{{\sin }^2}\,{{42}^ \circ } + 3{{\sin }^2}\,{{48}^ \circ }}}\)

According to the trigonometry properties 

⇒ {cos 29^o (1/sin 61^o) tan 45^o + 2 sin 35^o (1/cos 55^o)}/ 3sin² 42^o + 3sin²(90^o – 48^o)

⇒ {(cos 29^o/cos61^o)tan 45^o + 2 sin 35^o/cos 55^o}/3(sin 42^o + cos42^o)

⇒ {(cos29^o/sin(90 ^o – 61^o))× 1 + 2 sin 35^o/cos(90^o – 55^o)}/3 × 1

⇒ {cos 29²/cos 29² + 2 sin 35^o/sin 35^o}/3

⇒ (1 + 2)/3

⇒ 1

∴ The value of \(\frac{{\cos \,{{29}^ \circ }\cos ec\,{{61}^ \circ }\tan \,{{45}^ \circ } + 2\sin \,{{35}^ \circ }\sec \,{{55}^ \circ }}}{{3{{\sin }^2}\,{{42}^ \circ } + 3{{\sin }^2}\,{{48}^ \circ }}}\) is 1.

Given:

It is given that sinA = 3/5

Formula Used:

Basic concept of trigonometric ratio and identities

We know that

Sinθ = Perpendicular/Hypotenuse

Cosθ = Base/ Hypotenuse

Tanθ = Perpendicular/Base

(Hypotenuse)² = (Perpendicular)² + (Base)²

Calculation:

∵ sinA = 3/5

∴ P = 3 and h = 5

Now, by the help of Pythagoras theorem

∴ B² = 25 – 9 = 16

⇒ B = 4

Now we have to find the value of tanA + cosA

∴ tanA + cosA = P/B + B/H = 3/4 + 4/5 = 31/20

Hence, option (1) is correct

GIven:

a SinA + b cosA = c

Formula used:

If a Sinθ + b Cosθ = c

Then, a Cosθ – b Sinθ = \(\sqrt {{a^2} + {b^2} - {c^2}}\)

Calculation:

Apply above formula on a SinA + b cosA = c 

We get

⇒ a CosA – b SinA = \(\sqrt {{a^2} + {b^2} - {c^2}}\)

∴ a CosA – b SinA is equla to  \(\sqrt {{a^2} + {b^2} - {c^2}}\)

GIVEN:

tan X = 2; tan Y = 4, tan Z =?

FORMULA:

If XYZ is triangle, then, X + Y + Z = 180°

tan X + tan Y + tan Z = tan X tan Y tan Z

CALCULATION:

tan X + tan Y + tan Z = tan X tan Y tan Z

⇒ 2 + 4 + tan Z = 2 × 4 × tan Z

⇒ 6 + tan Z = 8tanZ

⇒ 7tan Z = 6

⇒ tan Z = 6/7

Ans: 4

Formula used:

(hypotenuse)² = (base)² + (height)²

Calculation:

This question will be solved by taking option

First we pick option 1

⇒ 5² + 12² = 25 + 144 = 169 = 13²

∴ option 1 satisfies condition of right angle triangle.

Now we take option 2

⇒ 3² + 4² = 9 + 16 = 25 = 5²

∴ option 2 also  satisfies condition of right angle triangle.

Now we take option 3

⇒ 6² + 8² = 36 + 64 = 100 = 10²

∴ option 3 also satisfies condition of right angle triangle.

Now we take option 4

⇒ 5² + 7² = 25 + 49 = 74 ≠ 9²

∴ values of option 4 are not satisfying condition of right angle triangle. 

GIVEN:

\(\frac{{sinθ [\left( {1 - tanθ } \right)tanθ + {{\sec }^2}θ ]}}{{\left( {1 - \sin θ } \right)\tan θ \left( {1 + tanθ } \right)\left( {secθ + tanθ } \right)}}\)

FORMULA USED:

\({\sec ^2}θ - \;{\tan ^2}θ = 1\), sin²θ + cos²θ = 1. tanθ = sinθ/cosθ, secθ = 1/cosθ 

CALCULATION:

\(\frac{{sinθ [\left( {1 - tanθ } \right)tanθ + {{\sec }^2}θ ]}}{{\left( {1 - \sin θ } \right)\tan θ \left( {1 + tanθ } \right)\left( {secθ + tanθ } \right)}}\)

\( ⇒ \;\frac{{\sin θ \;[\tan θ - \;{{\tan }^2}θ + \;1 + \;{{\tan }^2}θ ]}}{{(1 - \;\sin θ )\;\tan θ \;(1 + \;\tan θ )\;(\sec θ + \;\tan θ )}}\)

\( ⇒ \frac{{\sin θ \;(1 + \;\tan θ )}}{{(1 - \sin θ )\;\tan θ \;(1 + \;\tan θ )\;(\sec θ + \;\tan θ )}}\)

\(⇒ \;\frac{{\sin θ }}{{(1 - \;\sin θ )\;\frac{{\sin θ }}{{\cos θ }}\;(\sec θ + \;\tan θ )}}\)

\(⇒ \;\frac{{\cos θ }}{{(1 - \;\sin θ )\;\left( {\frac{1}{{\cos θ }} + \;\frac{{\sin θ }}{{\cos θ }}} \right)}}\)

\( ⇒ \frac{{{{\cos }^2}θ }}{{(1 - \sin θ )\;(1 + \;\sin θ )}}\)

\(⇒ \frac{{{{\cos }^2}θ }}{{(1 - \;{{\sin }^2}θ )}}\)

\(⇒ \;\frac{{{{\cos }^2}θ }}{{{{\cos }^2}θ }}\)

⇒ 1

As, tan²θ = sec²θ - 1

so, tanθ = \(\sqrt{sec^2θ-1}\) = \(\sqrt{x^2-1}\)

Given-

secθ = a/b

Concept Used- 

secθ = H/B, tanθ = P/B, sinθ = P/H and H² = P² + B² [where H = hypotenuse, B = base and P = perpendicular] 

Calculation-

According to Question -

H/B = a/b

P = √(a² - b²), H = a and B = b

tan²θ = (a2 - b2)/b²

sin²θ = (a2 - b2)/a2

(1 - tan²θ)/(2 - sin²θ)

⇒ {(2b² - a²)/b²}/{(a² + b²)/a²}

\(\Rightarrow \frac{{{a^2}\left( {2{b^2} - {a^2}} \right)}}{{{b^2}\left( {{a^2} + {b^2}} \right)}}\)

∴\(\frac{{1 - {{\tan }^2}\theta }}{{2 - {{\sin }^2}\theta }}=\frac{{{a^2}\left( {2{b^2} - {a^2}} \right)}}{{{b^2}\left( {{a^2} + {b^2}} \right)}}\)

Given:

5cos20° – 4sin230° + 6cosec245° = ?

Concept used:

using the concept of trigonometric ratios

Calculation:

5cos20° – 4sin230° + 6cosec245° = ?

⇒ 5 × (1)² – 4 × (1/2)² + 6 × (√2)² = ?

⇒ 5 – 4 × 1/4 + 6 × 2 = ?

⇒ 5 – 1 + 12 = ?

⇒ 4 + 12 = ?

⇒ ? = 16

∴ The value of ? is 16. 

Given:

Cotx = 5/12

Formula used:

Using basic trigonometric functions.

1) Cotx = B/P

2) Sinx = P/H

3) Cosx = B/H

Where B is base, P is perpendicular and H is the hypotenuse

Calculation:

Cotx = 5/12 = B/P

H² = B² + P² 

⇒ H² = 5² + 12² = 25 + 144

⇒ H² = 169

⇒ H = 13

Sinx + Cosx = P/H + B/H = 12/13 + 5/13

∴ Sinx + Cosx is 17/13

We know that values of cos x repeats after an interval of 2π or 360°.
Therefore, cos (–1710°) = cos (–1710° + 5 x 360°)
= cos (–1710° + 1800°) = cos 90° = 0.

Given:

The given condition is tan 4θ = cot (θ - 2)

Formula Used:

Basic concept of trigonometric ratio and identities

We know that

tan θ = cot (90 – θ)

Calculation:

∵ Cot θ = tan (90 – θ)

∴ cot (90 – 4θ) = cot (θ - 2)

⇒ (90 – 4θ) = (θ - 2)

⇒ θ = 18.4°

Now, we have to find the value of 7θ = 7 × 18.4° = 128.8°

Hence, option (4) is correct

Concept:

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

Calculation:

Let x = tan \(\pi\over8\)

x = tan (\({\pi\over4} - {\pi\over8}\))

x = \(\tan {\pi\over4} - \tan {\pi\over8}\over1+\tan {\pi\over4}\times\tan {\pi\over8}\)

x = \(\rm 1 - x\over 1+ x\)

x(1 + x) = 1 - x

x² + 2x - 1 = 0

\(\rm x = {-2 \pm \sqrt{2^2-4(-1)} \over 2}\)

\(\rm x = {-2 \pm 2\sqrt2 \over 2}\)

x = -1 + √2 or -1 - √2

Negative value cannot be possible in first quadrant 

∴ tan \(\pi\over8\) = x = √2 - 1

Given:

tanx/tany = a

Formula used:

sin (x + y) = sinx.cosy + cosx.siny

sin (x - y) = sinx.cosy - cosx.siny

tan θ = sin θ/cos θ 

Calculation:

∵ tanx/tany = a

⇒ sinx.cosy/cosx.siny = a

⇒ sinx.cosy = a × (cosx.siny)      ------(1)

∵ sin (x + y)/sin (x - y) = (sinx.cosy + cosx.siny)/(sinx.cosy - cosx.siny)

⇒ [a × (cosx.siny) + cosx.siny]/[a × (cosx.siny) - cosx.siny]

⇒ [cosx.siny(a + 1)]/[cos.siny(a - 1)]

⇒ (a + 1)/(a - 1)

Concept:

cos-1 x = \(π\over2\) - sin-1​ x

cot-1 x = \(π\over2\) - tan-1​ x

sec^-1 x = \(π\over2\) - cosec^-1 x

Calculation:

Given 3tan-1 x + cot-1 x = π

∵ cot-1 x = \(π\over2\) - tan-1​ x

3tan-1 x + (\(π\over2\) - tan-1​ x) = π

2tan^-1 x = \(\pi\over2\)

tan^-1 x = \(\pi\over4\)

x = tan \(\pi\over4\) = 1

Concept

Sec²A – Tan²A = 1

Cosec²A – Cot²A = 1

Cot(90° – A) = TanA

Calculation

Sec275° – Cot215° + Cosec265° – Cot265°

Sec² 75° – Cot²(90° – 75°) + 1

⇒  Sec²75° – Tan² 75° + 1

⇒  1 + 1

∴ 2

Given

cosec A + sec A = 2√2

Concept

cosec 45° = √2

sec 45° = √2

Calculation

cosec A + sec A = 2√2

L.H.S

cosec 45 °+ sec 45°

⇒  √2 + √2

⇒  2√2

So, A is 45°

Now, sin 45° = 1/√2

Calculation

(1 + Cos²A – 2 Cos A) + (1 + tan²A – 2 tan A) + (1 + sin²A – 2 Sin A)/ [2 – (cos A + tan A + sin A) + Tan² A]

⇒  (1 + 1 + 1 + Cos²A + Sin²A – 2cos A – 2 tan A – 2 sin A + Tan² A)/ [2 – (cos A + tan A + sin A) + Tan² A

⇒  [1 + 1 + 1 + 1 – 2(cos A + tan A + sin A) + Tan² A]/[2 – (cos A + tan A + sin A) + Tan² A]

⇒  {[4 – 2(cos A + tan A + sin A) ]+ Tan² A}/ [2 – (cos A + tan A + sin A) + Tan² A]

⇒  2{[2 – (cos A + tan A + sin A) ]+ Tan ² A}/[2 – (cos A + tan A + sin A) + Tan ² A]