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$(\frac{{\COS \theta }}{{1 - \sin \theta }} - \frac{{\cos \theta }}{{1 + sin\theta }}\; = \;\SQRT {12} \Rightarrow \;\frac{{\cos \theta (1 + \sin \theta ) - \cos \theta (1 - \sin \theta )}}{{1 - {{\sin }^2}\theta }}\; = \;2\sqrt 3 )$

(? (a + b)(a – b) = a² – b² & sin² x + cos² x = 1)

⇒ cosθ + cosθsin θ – cosθ + cosθsinθ = 2√3cos²θ

⇒ 2sinθcosθ = 2√3cos²θ

⇒ sinθ/cosθ = √3

(? tan x = sin x/cos x)

⇒ tan θ = √3

⇒ θ = 60° = π/3

<P>Given, ∠Q = 90, tan P = 24/7 and PQ = 14 cm

tan P = perpendicular/base = QR/PQ = 24/7

⇒ 24/7 = QR/PQ

⇒ QR = 24/7 × 14 = 48 cm

$({\RM{Given\;Sin\;}}10 \times {\rm{\;Sin\;}}30 \times {\rm{Sin\;}}50 \times {\rm{sin\;}}70)$

$(\Rightarrow \left( {\FRAC{1}{2}} \right) \times {\rm{Sin\;}}10 \times {\rm{Sin\;}}50 \times {\rm{sin\;}}70)$ (multiply and divide with 2 cos 10)

$(\Rightarrow \frac{{{\rm{\;}}2 \times {\rm{cos}}10 \times {\rm{Sin\;}}10 \times {\rm{Sin\;}}50 \times {\rm{\;sin\;}}70{\rm{\;}}}}{{\left( {2 \times 2 \times {\rm{cos}}10} \right)}})$ ? (2SINA cosA = Sin2A)

$(\Rightarrow \frac{{{\rm{\;}}\left( {{\rm{\;Sin\;}}20{\rm{\;}} \times {\rm{Sin\;}}50 \times {\rm{sin\;}}70} \right)}}{{{\rm{\;}}4{\rm{\;cos}}10}})$ (Multiply and divide with 2)

$(\Rightarrow \frac{{2 \times {\rm{Sin\;}}20 \times {\rm{\;Sin\;}}50 \times {\rm{sin\;}}\left( {90{\rm{\;}} - {\rm{\;}}20} \right)}}{{2 \times {\rm{\;}}4{\rm{\;cos}}10}} \Rightarrow {\rm{\;}}\frac{{2 \times {\rm{Sin\;}}20 \times {\rm{cos}}20 \times {\rm{Sin\;}}50}}{{8{\rm{\;cos}}10}})$

$(\Rightarrow \frac{{{\rm{Sin\;}}40{\rm{\;}} \times {\rm{Sin\;}}50}}{{8{\rm{\;cos}}10}})$ Again multiply and divide with 2.

$(\Rightarrow \frac{{\left( {2 \times \;Sin\;40\; \times Sin\;\left( {\;90\; - \;40} \right)} \right)\;}}{{2\;x\;8\;cos10}})$

$(\Rightarrow \frac{{\;2 \times Sin\;40\; \times cos40}}{{2\;x\;8\;cos\left( {\;90\; - \;80} \right)}})$

Given,

tan θ = 1/√7

? tan X = Opposite side/Adjacent side

Opposite side = 1 unit

Adjacent side = √7 units

By Pythagorean Theorem,

(Hypotenuse)² = (Side)² + (Side)²

⇒ Hypotenuse² = 1² + (√7)² = 8

⇒ Hypotenuse = √8 = 2√2

$(\BEGIN{array}{L} \frac{{co{t^2}\theta - {{\sec }^2}\theta }}{{SE{c^2}\theta + {{\cot }^2}\theta }}\; = \;\frac{{7 - \frac{8}{7}}}{{\frac{8}{7} + 7}}\; = \;\frac{{41}}{{57}}\\ \therefore \frac{{co{t^2}\theta - {{\sec }^2}\theta }}{{se{c^2}\theta+{{\cot }^2}\theta}}=\frac{{41}}{{57}}\END{array})$

$(\begin{array}{l} \Rightarrow \frac{{\LEFT( {\sin \THETA + \cos \theta } \right)\left( {\tan \theta + \cot \theta } \right)}}{{\sec \theta + {\rm{COSEC}}\theta }}\\ \Rightarrow \frac{{\left( {\sin \theta + \cos \theta } \right)\left( {\frac{{sin\theta }}{{cos\theta }} + \frac{{cos\theta }}{{sin\theta }}} \right)}}{{\frac{1}{{cos\theta }} + \frac{1}{{sin\theta }}}}\\ \Rightarrow \frac{{\frac{{\left( {\sin \theta + \cos \theta } \right)\left( {(SI{n^2}\theta + co{s^2}\theta } \right)}}{{sin\theta cos\theta }})}}{{\frac{{sin\theta + cos\theta }}{{sin\theta cos\theta }}}} \end{array})$

? sin² θ + cos² θ = 1

$(\frac{{9{\rm{sin\theta \;}} + {\rm{\;}}7{\rm{cos\theta \;}}}}{{{\rm{CO}}{{\rm{s}}^3}{\rm{\theta \;}} + {\rm{\;}}3{\rm{si}}{{\rm{n}}^3}{\rm{\theta \;}} + {\rm{\;}}5{\rm{sin\theta \;}}}})$

DIVIDING the expression by sinθ

$(\Rightarrow {\rm{\;}}\frac{{9{\rm{\;}} + {\rm{\;}}7{\rm{cot\theta \;}}}}{{{\rm{cot\theta \;co}}{{\rm{s}}^2}{\rm{\theta \;}} + {\rm{\;}}3{\rm{si}}{{\rm{n}}^2}{\rm{\theta \;}} + {\rm{\;}}5{\rm{\;}}}})$ 

$(\Rightarrow {\rm{\;}}\frac{{9{\rm{\;}} + {\rm{\;}}7{\rm{cot\theta \;}}}}{{{\rm{cot\theta \;co}}{{\rm{s}}^2}{\rm{\theta \;}} + {\rm{\;}}3{\rm{si}}{{\rm{n}}^2}{\rm{\theta \;}} + {\rm{\;}}5{\rm{\;}}}})$ 

Putting cotθ = 3

$(\Rightarrow {\rm{\;}}\frac{{30{\rm{\;}}}}{{3{\rm{co}}{{\rm{s}}^2}{\rm{\theta \;}} + {\rm{\;}}3{\rm{si}}{{\rm{n}}^2}{\rm{\theta \;}} + {\rm{\;}}5{\rm{\;}}}})$ 

$(\Rightarrow {\rm{\;}}\frac{{30{\rm{\;}}}}{{3({\rm{co}}{{\rm{s}}^2}{\rm{\theta \;}} + {\rm{\;si}}{{\rm{n}}^2}{\rm{\theta }}){\rm{\;}} + {\rm{\;}}5{\rm{\;}}}}{\rm{}} = {\rm{}}\frac{{15}}{4})$ 

$(\therefore {\rm{\;}}\frac{{9{\rm{sin\theta \;}} + {\rm{\;}}7{\rm{cos\theta \;}}}}{{{\rm{co}}{{\rm{s}}^3}{\rm{\theta \;}} + {\rm{\;}}3{\rm{si}}{{\rm{n}}^3}{\rm{\theta \;}} + {\rm{\;}}5{\rm{sin\theta \;}}}}{\rm{}} = {\rm{}}\frac{{15}}{4})$ 

According to formula,

MAXIMUM value of (a sin x ± b cos x) = √(a² + b²)

Minimum value of (a sin x ± b cos x) = - √(a² + b²)

∴ Maximum value of (9sinθ + 13 cosθ) is

sin x + sin² x = 1----(1)

cos² x + cos² x = ?----(2)

From equation (1)

sin x + sin² x = 1 (sin²x + cos²x = 1)

sin x + 1 – cos² x = 1

sin x = cos² x

sin² x = cos⁴ x

substituting VALUE of cos² x and cos⁴ x in (2)

We know that,

sec²θ - tan²θ = 1

Given, xcos θ – sinθ = 1

⇒ xcosθ = 1 + sinθ

Dividing the above EXPRESSION by cosθ

⇒ X = secθ + tanθ

Squaring both SIDES, we get

⇒ x² = (secθ + tanθ)²

⇒ x² = sec²θ + tan²θ + 2secθ tanθ

⇒ x² = 1 + tan²θ + tan²θ + 2secθ tanθ

⇒ x² = 1 + 2 tan²θ + 2secθ tanθ . . . . . . . eq (1)

Adding 1 on both sides, we get

⇒ 1 + x² = 2 + 2tan²θ + 2secθtanθ

⇒ (1 + x²) sinθ = 2sinθ + 2tan²θsinθ + 2secθtanθsinθ

⇒ (1 + x²) sinθ = 2sinθ + 2tan²θsinθ + 2tan²θ

⇒ x² + (1 + x²)sinθ = 1 + 2 tan²θ + 2secθ tanθ + 2sinθ + 2tan²θ sinθ + 2tan²θ (from eq (1))

⇒ x² + (1 + x²)sinθ = 4 tan²θ + 1 + 2sinθ + 2tan²θ sinθ + 2secθ tanθ

Using sec²θ – 1 = tan²θ, tanθ = sinθ/cosθ and secθ = 1/cosθ

 ⇒ x² + (1 + x²)sinθ = 4sec²θ -4 + 1 + 2sinθ + 2 sec²θsinθ – 2sinθ + 2sec²θsinθ

⇒ x² + (1 + x²)sinθ = 4sec²θ (1 + sinθ) - 3

Given,

$({\RM{cosec}}\THETA= \frac{{{x^2} + {y^2}}}{{{x^2} - {y^2}}})$ 

As,

cot²θ = cosec²θ – 1

$(\begin{array}{l} \Rightarrow co{t^2}\theta= \;{\LEFT( {\frac{{{x^2} + {y^2}}}{{{x^2} - {y^2}}}} \right)^2} - 1\\ \Rightarrow co{t^2}\theta= \;\frac{{{x^4} + {y^4} + 2{x^2}{y^2}}}{{{x^4} + {y^4} - 2{x^2}{y^2}}} - 1\\ \Rightarrow co{t^2}\theta= \;\frac{{{x^4} + {y^4} + 2{x^2}{y^2} - {x^4} - {y^4} + 2{x^2}{y^2}}}{{{x^4} + {y^4} - 2{x^2}{y^2}}}\\ \Rightarrow co{t^2}\theta= \;\frac{{4{x^2}{y^2}}}{{{{\left( {{x^2} - {y^2}} \right)}^2}}}\\ \Rightarrow \cot \theta= \;\frac{{2xy}}{{{x^2} - {y^2}}}\end{array})$

$(\begin{array}{l} \RIGHTARROW {\rm{\;}}x\; = \;secA - tanA\\ \Rightarrow {\rm{\;}}x\; = \;\frac{1}{{cosA}} - \frac{{sinA}}{{cosA}}\\ \Rightarrow {\rm{\;}}x\; = \;\frac{{1 - sinA}}{{cosA}}\; = \;\frac{{1 - sinA}}{{\SQRT {1 - {{\SIN }^2}A} }}\; = \;\frac{{\sqrt {{{\left( {1 - sinA} \RIGHT)}^2}} }}{{\sqrt {1 - {{\sin }^2}A} }}\\ \Rightarrow {\rm{\;}}x\; = \;\sqrt {\frac{{{{\left( {1 - sinA} \right)}^2}}}{{{1^2} - {{\sin }^2}A}}} \; = \;\sqrt {\frac{{\left( {1 - sinA} \right) \times \left( {1 - sinA} \right)}}{{\left( {1 - sinA} \right) \times \left( {1 + sinA} \right)}}} \\ \therefore {\rm{\;}}x\; = \;\sqrt {\frac{{1 - sinA}}{{\;1 + sinA}}} \end{array})$

tan θ + COT θ = x

x = (sinθ/cosθ) + (cosθ/sinθ) = (sin²θ + cos²θ)/sinθ cosθ = 1/sinθ cosθ

cos θ + sin θ = y

y² = cos²θ + sin²θ + 2sinθ cosθ

y² – 1 = 2 sinθ cosθ

COS 4a = sin 14a

⇒ cos 4a = cos (90 – 14a)

⇒ 4a + 14a = 90°

⇒ a = 5°

tan 6A + cot 6a = tan 6(5) + cot 6(5)

⇒ tan 30° + cot 30°

$(\RIGHTARROW \frac{1}{{\sqrt 3 }} + \sqrt 3)$

$(\Rightarrow \frac{4}{{\sqrt 3 }} \times \frac{{\sqrt 3 }}{{\sqrt 3 }})$

[(sin59°cos31° + cos59°sin31°) ÷ (cos20°cos25° - sin20°SIN 25°)]

APPLYING the FORMULA,

Sin(A + B) = SinACosB + CosASinB

Cos(A + B) = COSACOSB - SinASinB

∴ [(sin59°cos31° + cos59°sin31°) ÷ (cos20°cos25° - sin20°sin 25°)] = Sin90°/Cos45° = √2

$\sin x = \frac{3}{5} $

$\tan x = \frac{3}{4} $

$\SEC x = \frac{5}{4} $

$tan x + sec x = \frac{3}{4} + \frac{5}{4} = 2 $

 From GIVEN DATA,

⇒ COT ( – 5π/4) = X

⇒ cot (-225°) = x

⇒ - cot (180° + 45°) = x

⇒ - cot 45° = x

⇒ - 1 = x

We KNOW that, tan 60° = (√3).

So, θ - 30° = 60

So, θ = 90°.

SINA = cosA when θ is minimum

∴ A = 45°$

tanA - COTA = tan45° - cot45°

∴ 1 - 1 = 0

⇒ cosA + COSB = 2COS[(A + B)/2].COS[(A - B)/2]

APPLYING the formula on (cos3θ + cos7θ) in the numerator and on (cosθ + cos5θ) in denominator, we get

⇒ (2cos5θ.cos2θ + 2cos5θ) / (2cos3θ.cos2θ + 2cos3θ) + sin2θ.tan3θ

⇒ [{2cos5θ × (cos2θ + 1)} / {2cos3θ × (cos2θ + 1)}] + (sin2θ.sin3θ)/cos3θ

⇒ (cos5θ/cos3θ) + {(sin2θ.sin3θ)/cos3θ}

⇒ (cos5θ + sin2θ.sin3θ) / cos3θ

⇒ [cos(3θ + 2θ) + sin2θ sin3θ]/cos3θ

⇒ [cos3θ.cos2θ - sin2θ.sin3θ + sin2θ.sin3θ]/cos3θ = cos2θ

$(\because 1 - {\sin ^2}x = \LEFT( {1 - \sin x} \right)\left( {1 + \sin x} \right))$

$(\FRAC{{\left( { - 1} \right)\left( {1{\RM{\;}} + {\rm{\;sinx}}} \right)\left( {1 - \sin {\rm{x}}} \right)}}{{\sin {\rm{x\;}} + {\rm{\;}}1}} = \sin {\rm{x}} - 1)$

$(\begin{ARRAY}{l} \Rightarrow \;\sqrt {\frac{{1 + SIN\theta }}{{1 - sin\theta }}} \; = \;\sqrt {\frac{{1 + sin\theta }}{{1 - sin\theta }}\; \TIMES \;\frac{{1 + sin\theta }}{{1 + sin\theta }}} \; = \;\sqrt {\frac{{{{\left( {1 + sin\theta } \right)}^2}}}{{1 - {{\sin }^2}\theta }}} \; = \;\frac{{1 + sin\theta }}{{cos\theta }}\; = \;\frac{1}{{cos\theta }} + \frac{{sin\theta }}{{cos\theta }}\; = \;sec\theta+ tan\theta \\\THEREFORE \;\sqrt {\frac{{1 + sin\theta }}{{1 - sin\theta }}} \; = \;sec\theta+ tan\theta\end{array})$

sec²33° + sin²27° – cot²57° + sin²63°

= sec²33° – cot²(90° – 33°) + sin²27° + sin²(90° – 27°)

(? cot (90 – X) = tan x & sin (90 – x) = cos x)

= sec²33° – tan²33° + sin²27° + cos²27°

(? sin²x + cos²x = 1 &sec²x – tan²x = 1)

= 1 + 1 = 2

TAN(90° - θ) = cotθ

Sin(90° - θ) = cosθ

Sin²θ + cos²θ = 1

Sec²θ – tan²θ = 1

cosec²θ – COT²θ = 1

Given,

$(\frac{{{{\sec }^2}\theta- {{\cot }^2}\left( {90^\circ- \theta } \right)}}{{COSE{c^2}67^\circ- {{\tan }^2}23^\circ }} + {\sin ^2}40^\circ+ {\sin ^2}50^\circ)$

$(= \frac{{{{\sec }^2}\theta- {{\tan }^2}\theta }}{{cose{c^2}67^\circ- {{\tan }^2}(90^\circ- \;67^\circ )}} + {\sin ^2}40^\circ+ {\sin ^2}\left( {90^\circ- 40^\circ } \right))$

$(= \frac{1}{{cose{c^2}67^\circ- {{\cot }^2}67^\circ }} + \;{\sin ^2}40^\circ+ co{s^2}40^\circ)$

= 1 + 1

$(\sqrt {\frac{{1 + {\RM{sin\THETA }}}}{{1 - {\rm{sin\theta }}}}} )$

$(= \sqrt {\frac{{1 + {\rm{sin\theta }}}}{{1 - {\rm{sin\theta }}}} \times \frac{{1 + {\rm{sin\theta }}}}{{1 + {\rm{sin\theta }}}}} )$

$(= \frac{{1 + {\rm{sin\theta }}}}{{\sqrt {1 - {\rm{si}}{{\rm{n}}^2}{\rm{\theta }}} }})$

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

 secθ + tanθ = 

⇒ COS 330°

⇒ cos (360° - 30°)

⇒ cos 30°(COSINE is POSITIVE in FOURTH QUADRANT)

From the GIVEN DATA,

Measure of an exterior angle of a polygon with n SIDES = 360°/n

For dodecagon has 12 sides, n = 12

Given, X = sin θ + cos θ and y = sin θ – cos θ

Thus, x + y = 2sinθ

Also, xy = sin²θ – cos²θ

(x + y)² = x² + y² + 2xy

⇒ 4sin²θ = x² + y² + 2sin²θ – 2cos²θ

⇒ x² + y² = 2(sin²θ + cos²θ) = 2

Since we know that tan 45 = 1 and sec 60 = 2,

Put this in the REQUIRED expression

⇒ Tan 45° + 4/√3 Sec 60° = 1 + 8/ √3

⇒ Tan 45° + 4/√3 Sec 60° = (√3 + 8)/√3

∴ Tan 45° + 4/√3 Sec 60° = (√3 + 8)/√3

Solution;

Given: Height, h = 75M which is the OPPOSITE side.

θ is the angle subtended by the string with the ground level and,

Length of the string = ? (This is the hypotenuse of the triangle).

It's also given that,

cotθ = 8/15

We know that, cotθ = opposite side / adjacent side 

So, adjacent side /opposite side = 8/15

adjacent side/75= 8/15

Adjacent side = (75×8)/15= 600/15= 40

Now, we know that 

Hypotenuse² = (opposite side)² + (adjacent side)²

So, Hypotenuse ² = 75² + (40)²

Hypotenuse ² =5625 + 1600

Hypotenuse ² = 7225

Hypotenuse = 85m 

The length of the string is 85m.

So, the correct option is 2).85m

cos θ = √(1 – sin²θ) and secθ = 1/cosθ

sin θ = 5/13

$(cos\theta= \sqrt {1 - {{\left( {\frac{5}{{13}}} \right)}^2}})$ 

⇒ cos θ = 12/13

sec θ = 13/12

SINCE we know that SEC 30 = 2/√3 and tan 60 = √3, put this in the required equation

⇒ (½) Sec 30° + √2 tan 60°

⇒ (1/2) × 2/√3 + (√2) × (√3) = (1 + 3√2)/√3

∴ the value of (1/2) Sec 30° + √2 Tan 60° is (1 + 3√2)/√3

⇒ 1/sin A + COSA/sinA = 3

⇒ cosA + 1 = 3SIN A

⇒ On squaring both sides, we get:-

⇒ COS²A + 2 cosA + 1 = 9 sin²A = 9(1 – cos²A)

⇒ 10cos²A + 2cosA – 8 = 0

⇒ 5cos²A + cosA – 4 = 0

⇒ 5cosA (cosA + 1) – 4 (cosA + 1) = 0

⇒ (cosA + 1) (5cosA – 4) = 0

2cos²θ + 11 sinθ - 7 = 0

2(1 - sin²θ) + 11 sinθ - 7 = 0

2sin²θ - 11 sinθ + 5 = 0

2sin² θ - 10 sin θ - sin θ + 5 = 0

2sinθ (sinθ - 5) - 1 (sinθ - 5) = 0

(2sinθ - 1) (sinθ - 5) = 0

∴ sinθ $( = \FRAC{1}{2})$

[As sinθ = 5 is not possible]

[As we know sinθ = 0 ≤ θ ≤ $(\frac{{\RM{\pi }}}{2})$] 

To SIMPLIFY the EXPRESSION: (cosec⁴A – cot²A) – (cot⁴A + cosec²A)

[IDENTITY USED: 1 + cot²A = cosec²A]

⇒ cosec⁴A – cot⁴A – cot²A – cosec²A

Since, cosec⁴A = 1 + cot⁴A + 2 cot²A, we have

⇒ 1 + cot⁴A + 2 cot²A – cot⁴A – cot²A – cosec²A

⇒ 1 + cot²A – cosec²A

⇒ cosec²A - cosec²A

⇒ 0

(sinθ + cosecθ) = 2.5

sinθ $(+\; \frac{1}{{{\rm{sin\theta }}}} = 2.5)$

-sin²θ + 2.5 sinθ - 1 = 0

sin²θ - 2.5sinθ + 1 = 0

sinθ $( = \frac{{ + 2.5 \PM \SQRT {6.25 - 4} }}{2})$

sinθ $( = \frac{{ + 2.5 \pm 1.5}}{2})$

∴ sinθ = 2,$(\frac{1}{2})$

sinθ = 2

∴ Not possible for 0 < θ < $(\frac{{\rm{\pi }}}{2})$

∴ sinθ = $(\frac{1}{2})$

∴ θ = 30°

$(\frac{{5{\rm{sin}}75^\circ {\rm{sin}}77^\circ + 2{\rm{cos}}13^\circ {\rm{cos}}15^\circ }}{{{\rm{cos}}15^\circ {\rm{sin}}77^\circ }} - \frac{{7{\rm{sin}}81^\circ }}{{{\rm{cos}}9^\circ }})$

$(= \frac{{{\rm{\;}}2\cos \left( {90^\circ - 75^\circ } \RIGHT){\rm{sin}}77^\circ + 2\sin \left( {90^\circ - 13^\circ } \right){\rm{cos}}15^\circ }}{{{\rm{cos}}15^\circ {\rm{sin}}77^\circ }} - 7\frac{{{\rm{sin}}81^\circ }}{{\sin \left( {90^\circ - 9^\circ } \right)}})$

$(= 5\frac{{{\rm{cos}}15^\circ {\rm{sin}}77^\circ }}{{{\rm{cos}}15^\circ {\rm{sin}}77^\circ }} + 2\frac{{{\rm{sin}}77^\circ {\rm{cos}}15^\circ }}{{{\rm{cos}}15^\circ {\rm{sin}}77^\circ }} - 7\frac{{{\rm{sin}}81^\circ }}{{{\rm{sin}}81^\circ }})$

= 5 + 2 - 7

= 7 - 7 = 0

Given, 5tanθ = 4

⇒tanθ = 4/5

We have to find the value of $(\LEFT( {\FRAC{{5\sin \theta - 3\cos \theta }}{{5\sin \theta + 3\cos \theta }}} \right))$

$(= \;\left( {\frac{{5\sin \theta - 3\cos \theta }}{{5\sin \theta + 3\cos \theta }}} \right))$

$(=\left( {\frac{{cos\theta (5\frac{{\sin \theta }}{{cos\theta }} - 3)}}{{cos\theta (5\frac{{\sin \theta }}{{cos\theta }} + 3)}}} \right))$ (taking cosθ COMMON from numerator and DENOMINATOR)

$(= \left( {\frac{{5tan\theta - 3}}{{5tan\theta + 3}}} \right))$

$(= \left( {\frac{{5 \times \frac{4}{5} - 3}}{{5 \times \frac{4}{5} + 3}}} \right))$

$(= \frac{{4 - 3}}{{4 + 3}})$

The addition FORMULA for the tangent may be used to write

⇒ [TAN(25^o) + tan(50^o)]/[1 - tan( 25^o) tan(50^o)] = tan(25^o + 50^o) = tan(75^o) = tan(45^o + 30^o)

⇒ [tan(45^o) + tan(30^o)]/[1 - tan(45^o)tan(30^o) ] = [ 1 + 1/√3 ]/[1 - 1 × 1/√3] = √3 + 2

We can write,

sec 30° + tan 60°

⇒ 1/cos 30° + SIN 60°/cos 60°(? cos 30° = sin 60° = √3/2, cos 60° = 1/2)

⇒ 2/√3 + (√3/2)/(1/2)

⇒ 2/√3 + √3

⇒ (2 + 3)/√3

⇒ 5/√3