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Given:

secx + tanx = 3

Concept Used:

If secx + tanx = y. then secx - tanx = 1/y

cosec²x - cot²x = 1

tanx = 1/cotx

Calculation:

secx + tanx = 3      ----(i)

secx - tanx = 1/3      ----(ii)

Subtracting (ii) from (i), we get 2tanx,

2tanx = 8/3

⇒ tanx = 4/3

⇒ cotx = 3/4

cosecx = √(1 + cot²x)

⇒ √(1 + 9/16)

⇒ √(25/16)

⇒ 5/4

cosecx + cotx = 5/4 +3/4 

⇒ 2

∴ The value of cosecx + cotx is 2.

Given:

sinx + cosx = √2cosx

Concept Used:

sin²x + cos²x = 1

Calculation:

sinx + cosx = √2cosx

Squaring the equation on both side, we get

⇒ sin²x + cos²x +2sinxcosx = 2cos²x

⇒ 1 + 2sinxcosx = 2cos²x      ----(i)

Let sinx - cosx = k      ----(ii)

Squaring the equation on both side, we get

⇒ 1 - 2sinxcosx = k²

Adding (i) and (ii), we get

2cos²x + k² = 2

⇒ k² = 2(1 - cos²x)

⇒ k² = 2sin²x

⇒ k = +√2sinx

∴ The value of sinx - cosx = +√2sinx

Given:

\(\cos θ = \frac 5 {13}\)

Concept used:

1 + tan2 θ = sec2 θ

cosθ = 1/secθ 

Calculations:

⇒ tan2 θ + sec2 θ

⇒ sec2 θ - 1 + sec2 θ

⇒ 2sec2 θ - 1

⇒ secθ = 13/5

⇒ (2 × 169)/25 - 1

⇒ (338/25) - 1

⇒ (338 - 25)/25

⇒ 313/25

∴ The value of tan2 θ + sec2 θ is 313/25.

Given:

 sec θ + tan θ = 3

Concept used:

sec θ + tan θ = x

sec θ - tan θ = 1/x

Calculations:

⇒  sec θ + tan θ = 3          ------(1)

⇒ sec θ + tan θ = 1/3          ------(2)

By adding equation (1) and (2)

⇒ 2secθ = 3 + 1/3

⇒ 2secθ = 10/3

⇒ secθ = 5/3

∴ The value of secθ is 5/3.

GIVEN:

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

CONCEPT USED:

sinθ₁ = cosθ₂ , when θ₁ + θ₂ = 90° 

CALCULATION:

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

⇒ (x + y) + (x - y) = 90° 

⇒ 2x = 90° 

⇒ x = 45° 

⇒ cos²x = cos²45° 

⇒ (1/√2)² 

⇒ 1/2

∴ cos2x  = 1/2

Given:

\(\frac 1 {\rm cosec^2\;51^\circ} + \sin^2 39^\circ + \tan^2 51^\circ - \frac 1 {\sin^2 51^\circ\sec^2 39^\circ}\)

Concept used:

sin²θ + cos²θ = 1

1 + cot²θ = cosec²θ 

sinθ = 1/x

sin(90° - θ) = cosθ 

sin39° = co51° = 1/x

Calculatuions:

⇒ sin²51° + sin²39° + tan²51° - (cos²39°/sin²51°)

⇒ sin251° + {sin(90° - 51°)}2 + tan251° - [cos239°/{sin(90° - 39°)}2]

⇒ sin251° + cos²51° + tan251° - [cos239°/cos²39°]                        

⇒ 1 + tan251° - 1

⇒ tan251°

⇒ {tan(90° - 39°)}²

⇒ cot²39° 

⇒ cosec²39° - 1

⇒ x² - 1

∴ The value is x2 - 1

Given:

cosec θ = 13/12

Forumula used:

H = hypotenuse, P = perpendicular, B = base

H² = P² + B²

cosecθ = H/P, sinθ = P/H, cosθ = B/P

Calculation:

⇒ cosec θ = 13/12 = H/P

⇒ H = 13, P = 12, B² = H² – P²

⇒ B² = 13² – 12² = 5²

⇒ B = 5

⇒ sinθ = 12/13, cosθ = 5/13, and cosec θ = 13/12

putting the value of sinθ and cosθ in \( \frac{{2sin\theta - 3cos\theta }}{{4sin\theta - 9cos\theta }}\)

⇒ \(\frac{{2 \times \frac{{12}}{{13}} - 3 \times \frac{5}{{13}}}}{{4 \times \frac{{12}}{{13}} - 9 \times \frac{5}{{13}}}}\)

⇒ 3

∴ The value of \( \frac{{2sin\theta - 3cos\theta }}{{4sin\theta - 9cos\theta }}\) is 3.

Given:

tanθ + cotθ = 2

Calculation:

Put the value of θ = 45^o, tan45^o = 1, and cot45^o = 1

Then, tanθ + cotθ = 2

⇒ tan45^o + cot45^o = 2

⇒ 1 + 1 = 2

putting the value θ = 45 in tan100θ + cot100θ

⇒ tan¹⁰⁰45 + cot¹⁰⁰45 

⇒ 1¹⁰⁰ + 1¹⁰⁰

⇒ 2

∴ The value of tan100θ + cot100θ is 2.

Given:

2cos²θ – 5cosθ + 2 = 0

Concept used:

Using the concept of trigonometric ratios with respect to standard angles.

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

cosθ ≠ 2, because maximum value of cosθ is 1.

So, cosθ = 1/2

⇒ cosθ = cos60°      [0° < θ < 90°]

⇒ θ = 60° 

\(\sqrt {\frac{{1{\rm{\;}} + {\rm{\;ta}}{{\rm{n}}^2}{\rm{θ }}}}{{1{\rm{\;}} - {\rm{\;co}}{{\rm{t}}^2}{\rm{θ }}}}} \)

\( \Rightarrow \sqrt {\frac{{1{\rm{\;}} + {\rm{\;ta}}{{\rm{n}}^2}60^\circ }}{{1{\rm{\;}} - {\rm{\;co}}{{\rm{t}}^2}60^\circ }}} \)

\( \Rightarrow \sqrt {\frac{{1{\rm{\;}} + {\rm{\;}}{{\left( {\sqrt 3 } \right)}^2}}}{{1{\rm{\;}} - {\rm{\;}}{{\left( {\frac{1}{{\sqrt 3 }}} \right)}^2}}}} \)

\( \Rightarrow \sqrt {\frac{{1{\rm{\;}} + {\rm{\;}}3}}{{1{\rm{\;}} - {\rm{\;}}\frac{1}{3}}}} \)

\( \Rightarrow \sqrt {\frac{4}{{{\rm{\;}}\frac{2}{3}}}} \)

⇒ √6

∴ The value is √6.

Given:

3 sec² x - 4 = 0

Formula:

sec 30° = 2/√3

Calculation:

3 sec² x - 4 = 0

⇒ sec² x = 4/3

⇒ sec x = 2/√3

⇒ sec x = sec 30°

∴ x = 30°

Given :

tan4A = cot(A – 20°)

Formula used :

tanθ = cot (90° – θ)

Calculations :

cot (90° – 4A) = cot (A – 20°)

⇒ 90°– 4A = A – 20° 

⇒ 5A = 110° 

⇒ A = 22° 

∴ The value of A will be 22° 

Given:

sinθ + cosecθ = 2

Formula Used:

cosecθ = 1/sinθ

Calculation:

Let sinθ = x

⇒ sinθ + cosecθ = x + 1/x = 2

⇒ (x + 1/x)² = 2²

⇒ x² + 2 + 1/x² = 4

⇒ x² + 1/x² = 2

∴ sin²θ + cosec²θ = 2

Given:

cos2(π/8) + 4cos2(π/4) – sec²(π/4) + 2tan2(π/3) + sin2(π/8)

Concept Used:

cos²θ + sin²θ = 1

π = 180° 

Calculation:

cos2(π/8) + 4cos2(π/4) – sec²(π/4) + 2tan2(π/3) + sin2(π/8)

⇒ {cos²(π/8) + sin²(π/8)} + 4cos²45° – sec²45° + 2tan²60°

⇒ 1 + 4(1/√2)² – (√2)² + 2(√3)²

⇒ 1 + 2 – 2 + 6

⇒ 7

∴ The value of the expression cos2(π/8) + 4cos2(π/4) – sec²(π/4) + 2tan2(π/3) + sin2(π/8) is 7.

Given:

cosec θ = α ⇒ cosec-1α = θ

secΦ = β ⇒ sec^-1β = Φ  

Formula used:

Sec x = cosec(90° – x)

Calculation:

Let sec^-1 γ  be x

⇒ sec x = γ 

⇒ cosec(90° – x) =  γ 

⇒ 90° – x = cosec^-1γ 

⇒ cosec-1​γ + x =  90°

⇒ cosec-1​γ + sec-1 γ = 90° 

∴  The value of cosec-1​γ + sec-1 γ is 90° 

Since cos x = - 3/5 , we have sec x = - 5/3
Now sin² x + cos² x = 1, i.e., sin² x = 1 – cos² x
or sin² x = 1 – 9/25 = 16/25
Hence sin x = ± 4/5
Since x lies in third quadrant, sin x is negative. Therefore
sin x = – 4/5
which also gives
cosec x = – 5/4

Further, we have
tan x = sinx/cosx = 4/3 and cot x = cosx/sinx = 3/4 .

Solution:

Let Ф = π/8
2Ф = π/4
tan 2Ф = 1  = 2 tan Ф / [ 1 - tan² Ф ]
1 - tan² Ф = 2 tan Ф
tan² Ф + 2 tan Ф - 1 = 0
tan Ф  =  1/2 [ -2 +- √(4 +4) ]  = -1 +- √2 
tan π/8 is > 0 

Therefore,  tan π/8 = √2 - 1

GIVEN:

\(\left[ {\frac{{\left( {sin\theta \; + \;cos\theta \; + \;1} \right)\left( {sin\theta \; + \;cos\theta - 1} \right)sec\theta \;cosec\theta }}{{\left( {co{s^6}\theta \; + \;si{n^6}\theta \; + \;3si{n^2}\theta \;co{s^2}\theta } \right)}}} \right]\)

FORMULA USED:

(a + b)(a- b) = a² – b²

\(co{s^6}\theta \; + \;si{n^6}\theta = 1 - 3{\sin ^2}\theta {\cos ^2}\theta \)

sin²θ + cos²θ = 1

CALCULATION:

\(\left[ {\frac{{\left( {sin\theta \; + \;cos\theta \; + \;1} \right)\left( {sin\theta \; + \;cos\theta - 1} \right)sec\theta \;cosec\theta }}{{\left( {co{s^6}\theta \; + \;si{n^6}\theta \; + \;3si{n^2}\theta \;co{s^2}\theta } \right)}}} \right]\)

⇒\(\left[ {\frac{{\left[ {{{\left( {sin\theta \; + \;cos\theta } \right)}^2} - 1} \right]sec\theta \;cosec\theta }}{{\left( {1 - 3{{\sin }^2}\theta {{\cos }^2}\theta \; + \;3si{n^2}\theta \;co{s^2}\theta } \right)}}} \right]\)

⇒\(\left[ {\frac{{({{\sin }^2}\theta \; + \;{{\cos }^2}\theta \; + \;2sin\theta \;cos\theta - 1)sec\theta \;cosec\theta }}{{\left( 1 \right)}}} \right]\)

⇒\(\left[ {\frac{{\left( {1\; + \;2sin\theta \;cos\theta - 1} \right)sec\theta \;cosec\theta }}{{\left( 1 \right)}}} \right]\)

⇒ 2sin θ cos θ sec θ cosec θ

⇒ 2

GIVEN:

\(\frac{{\left[ {4co{s^4}32^\circ cose{c^2}58^\circ - sin63^\circ cos27^\circ - co{s^2}63^\circ \; + \;1} \right]}}{{\left[ { - cose{c^2}24^\circ \; + \;ta{n^2}66^\circ - se{c^2}31^\circ \; + \;co{t^2}59^\circ } \right]{{\cos }^2}32^\circ }}\)

FORMULA USED:

sin(90° - x) = cos x, cosec(90° - x) = sec x

(1 – sin²x) = cos²x

(1 – cos²x) = sin²x

CALCULATION:

\(\frac{{\left[ {4co{s^4}32^\circ cose{c^2}58^\circ - sin63^\circ cos27^\circ - co{s^2}63^\circ \; + \;1} \right]}}{{\left[ { - cose{c^2}24^\circ \; + \;ta{n^2}66^\circ - se{c^2}31^\circ \; + \;co{t^2}59^\circ } \right]\;{{\cos }^2}32^\circ }}\)

= \(\frac{{\left[ {4co{s^4}32^\circ se{c^2}32^\circ - sin63^\circ sin63^\circ - co{s^2}63^\circ \; + \;1} \right]}}{{\left[ { - se{c^2}66^\circ \; + \;ta{n^2}66^\circ - cose{c^2}59^\circ \; + \;co{t^2}59^\circ } \right]\;{{\cos }^2}32^\circ }}\)

= \(\frac{{\left[ {4co{s^2}32^\circ - {{\sin }^2}63^\circ - co{s^2}63^\circ \; + \;1} \right]}}{{\left[ { - se{c^2}66^\circ \; + \;ta{n^2}66^\circ - cose{c^2}59^\circ \; + \;co{t^2}59^\circ } \right]\;{{\cos }^2}32^\circ }}\)

= \(\frac{{\left[ {4co{s^2}32^\circ } \right]}}{{\left[ { - 1 - 1} \right]{{\cos }^2}32^\circ }}\)

= 4/(- 2)

= - 2

It is known that the values of sin x repeat after an interval of 2π or 360°.

sin765^∘=sin(8×90^∘+45^∘)

sin⁡765^∘=sin⁡(8×90^∘+45^∘)

⇒sin45^∘=1/√2

Given - 

 ​​\(\frac{{sinθ }}{{1 + cosθ }}\; + \;\frac{{1 + cosθ }}{{sinθ }} = \frac{4}{{√ 3 }}\)

Concept used - 

sin²θ + cos²θ = 1

sinθ × sinθ = (1 - cos²θ)

sinθ × sinθ = (1 + cosθ) × (1 - cosθ)

{sinθ/(1 + cosθ)} = {(1 - cosθ)/sinθ}

Solution - 

\(\frac{{sinθ }}{{1 + cosθ }}\; + \;\frac{{1 + cosθ }}{{sinθ }} = \frac{4}{{√ 3 }}\)

\(⇒ \frac{{sinθ \: \times (1 - cosθ) }}{{(1 + cosθ)(1 -\ cosθ) }}\; + \;\frac{{1 + cosθ }}{{sinθ }} = \frac{4}{{√ 3 }}\)

⇒ Sinθ(1 - cosθ)/1 - cos²θ  + (1 + cosθ)/sinθ = 4/√3

⇒ Sinθ (1 - cosθ)/sin²θ + (1 + cosθ)/sinθ = 4/√3

\(⇒ \frac{{(1 - cosθ) }}{{sinθ }}\; + \;\frac{{1 + cosθ }}{{sinθ }} = \frac{4}{{√ 3 }}\)

⇒ (2/sinθ) = (4/√3)

⇒ sinθ = (√3/2) = sin60° 

⇒ θ = 60°

⇒ (secθ + cotθ + cosecθ) ^{- 1}

⇒ (sec60° + cot60° + cosec60°) ^{- 1}

⇒ {2 + (1/√3) + (2/√3)} ^{- 1}

⇒ (2 + √3)  ^{- 1}

⇒ 1/(2 + √3)

⇒ ( 2 - √3)

∴ Ans = (2 - √3).

Given:

\(\cot \theta = \frac{1}{{\sqrt 3 }}\)

θ = 60°

Calculation:

 sin 60° =\(\frac{{\sqrt 3 }}{2}\)  

 cos 60° = \(\frac{1}{2}\)

 cosec 60° =\(\frac{2}{\sqrt3}\)

 sec 60° = 2 

substituting the value in the given expression  

 = \(\frac{2-\frac{3}{4}\ }{1-\frac{1}{4}}+(\frac{4}{3}\ -2)\) ⇒\(\frac{\frac{5}{4}\ }{\frac{3}{4}}-\frac{2}{3}\)

 = \(\frac{{5\;}}{3} - \frac{2}{3} = 1\)

∴ the value for the \(\frac{{2 - {{\sin }^2}\theta }}{{1 - {{\cos }^2}\theta }} + (cose{c^2}\theta - {\sec} \theta )\) = 1

Concept:

sin (2nπ + θ) = sin θ

sin (2nπ - θ) = -sin θ

sin (90 + θ) = cos θ

Calculation:

To Find: Value of sin (2190°)

⇒ sin (2190°) = sin(360° × 6 + 30°) 

= sin (12π + 30°)             

= sin (30°)                    (∵ sin (2nπ + θ) = sin θ)

Given:

4(cosec2 57 - tan2 33) - cos 90 + y × tan2 66 × tan2 24 = y/2

Formula used:

(i) cosec(90 - θ) = secθ

(ii) tan(90 - θ) = cotθ

(iii) sec2θ - tan2θ = 1

(iv) tanθ = 1/cotθ 

Calculations:

4(cosec2 (90 - 33) - tan2 33) - cos 90 + y × tan2 66 × tan2 (90 - 66) = y/2

⇒ 4(sec2 57 - tan2 33) - cos 90 + y × tan2 66 × cot2 66 = y/2

⇒ 4 × 1 - 0 + y × tan2 66 × 1/tan2 66 = y/2

⇒ 4 + y = y/2

⇒ y - y/2 = -4

⇒ y/2 = -4

⇒ y = -8

∴ The value of y is -8.

Concept:

The domain of inverse sine function, sin x is \(x \in \left[ { - 1,1} \right]\)

Calculation:

Domain of the function is calculated as follows:

\({\sin ^{ - 1}}\left( {2x\sqrt {1 - {x^2}} } \right)\)

\(- 1 \le 2x\sqrt {1 - {x^2}} \le 1\)

\( - \frac{1}{2} \le x\sqrt {1 - {x^2}} \le \frac{1}{2}\)

\({x^2}\left( {1 - {x^2}} \right) \le \frac{1}{4}\)

\(t - {t^2} - \frac{1}{4} \le 0\)

\({\left( {t - \frac{1}{2}} \right)^2} \le 0\)

\(t \le \frac{1}{2}\)

\({x^2} \le \frac{1}{{\sqrt 2 }}\)

\(x \in \left[ { - \frac{1}{{\sqrt 2 }},\frac{1}{{\sqrt 2 }}} \right]\)

Angles are 45°, 60° and 75°. 

The ratio of smallest and greatest sides = sin45°; sin75°= √3 1:1

Solution: In 60 minutes, the minute hand of a watch completes one revolution. Therefore,
in 40 minutes, the minute hand turns through 2/3 of a revolution. Therefore,
θ = 2 × 360°/3
or 4π/3
radian. Hence, the required distance travelled is given by
l = r θ = 1.5 × 4π/3 cm = 2π cm = 2 × 3.14 cm = 6.28 cm.