Question

if cot theta =3/4 prove that √sec thetha -cosec thatha / sec thetha + cosec thetha = 1/√7​

Answer 1

EXPLANATION.

→ Cot ø = 3/4.

prove = √ sec ø - Csc ø / Sec ø + Csc ø.

→ Cot ø = B/P = Base/Perpendicular.

→ Cot ø = B/P = 3/4.

By using Pythagorean theorem,we get.

→ H² = P² + B².

→ H² = (4)² + (3)²

→ H² = 16 + 9.

→ H² = 25

→ H = √25 = 5.

→ Sin ø = P/H = Perpendicular/Hypotenuse

Sin ø = 4/5.

→ Cos ø = B/H = Base/Hypotenuse = 3/5.

→ Tan ø = P/B = Perpendicular/Base = 4/3.

→ Csc ø = H/P = Hypotenuse/Perpendicular

Csc ø = 5/4.

→ Sec ø = H/B = Hypotenuse/Base = 5/3.

→ Cot ø = B/P = Base/Perpendicular = 3/4.

$\sf \: \implies \: \sqrt{ \dfrac{ \sec( \theta) - \csc( \theta) }{ \sec( \theta) + \csc( \theta) } } = \dfrac{1}{ \sqrt{7} }$

$\sf \: \implies \: \sqrt{ \dfrac{ \dfrac{5}{3} \: - \: \dfrac{5}{4} }{ \dfrac{5}{3} \: + \: \dfrac{5}{4} } } = \sqrt{ \dfrac{ \dfrac{20 - 15}{12} }{ \dfrac{20 + 15}{12} } } \\ \\ \\ \sf \: \implies \: \sqrt{ \dfrac{ \dfrac{5}{12} }{ \dfrac{35}{12} } } \: = \: \sqrt{ \frac{5}{12} } \times \sqrt{ \frac{12}{35} } \\ \\ \sf \: \implies \: \sqrt{ \frac{1}{7} } = proved$

Answer 2

EXPLANATION.

→ Cot ø = 3/4.

prove = √ sec ø - Csc ø / Sec ø + Csc ø.

→ Cot ø = B/P = Base/Perpendicular.

→ Cot ø = B/P = 3/4.

By using Pythagorean theorem,we get.

→ H² = P² + B².

→ H² = (4)² + (3)²

→ H² = 16 + 9.

→ H² = 25

→ H = √25 = 5.

→ Sin ø = P/H = Perpendicular/Hypotenuse

Sin ø = 4/5.

→ Cos ø = B/H = Base/Hypotenuse = 3/5.

→ Tan ø = P/B = Perpendicular/Base = 4/3.

→ Csc ø = H/P = Hypotenuse/Perpendicular

Csc ø = 5/4.

→ Sec ø = H/B = Hypotenuse/Base = 5/3.

→ Cot ø = B/P = Base/Perpendicular = 3/4.

√sec(θ)+csc(θ)sec(θ)−csc(θ)√1

√sec(θ)+csc(θ)sec(θ)−csc(θ) √7

⟹ √5/3 -5/4/5/4+5/4=√20-15/12/20+15/12

⟹ √5/12/35/12=√5/12×√12/13

⟹ √1/7