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1.

\(\frac{{sin\theta }}{{1 + cos\theta }}\; + \;\frac{{1 + cos\theta }}{{sin\theta }} = \frac{4}{{\sqrt 3 }}\); 0°<θ < 90°, then find the value of (secθ + cotθ + cosecθ)- 11. (2 -√3)2. (2 +√3)3. (√3 - 2)4. (√2 +√3)

Answer» Correct Answer - Option 1 : (2 -√3)

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).
Discussion
2.

If\(\cot \theta = \frac{1}{{\sqrt 3 }}\), 0° <θ° < 90°, then the value of\(\frac{{2 - {{\sin }^2}\theta }}{{1 - {{\cos }^2}\theta }} + (cose{c^2}\theta - {\sec} \theta )\) is:1. 12. 53. 04. 2

Answer» Correct Answer - Option 1 : 1

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
Discussion