Question

1) Show that, cot30° - cot75°/tan15° - tan60° = 1

2) If cotΘ = a/b, Find a secΘ - b cosecΘ​

Answer 1

Answer:

$\huge \boxed{ \fcolorbox{red}{pink}{answer}}$

➡Show that, cot30° - cot75°/tan15° - tan60° = 1.......

Answer ;

$as \: we \: know \: that \: \\ cot30 = \cot(90 - 60) \\ \cot(90 - 60) = tan60 \\ similarily \\ \cot(75) = \cot(90 - 15) = tan(15) \\ \frac{tan(60) - tan(15)}{tan(15) - tan(60)} \\ - 1 \: answer$

Hope I help you!!

Answer 2

Question 1) :- Show that, cot30° - cot75°/tan15° - tan60° = (-1) .

Solution :-

we know that,

• cotA = tan(90 - A)

So, changing both numerator value in tan, we get,

• cot30° = tan(90° - 30°) = tan60°
• cot75° = tan(90° - 75°) = tan15°

Putting both values now we get,

→ (tan60° - tan15°) / (tan15° - tan60°)

taking (-1) common from numerator now,

→ (-1)(tan15° - tan60°) / (tan15° - tan60°)

→ (-1) (Proved).

___________________

Question 2) :- If cotΘ = a/b, Find asecΘ - bcosecΘ

Solution :-

we know that ,

• cotA = Base/Perpendicular
• Base = a
• perpendicular = b

So,

→ Hypotenuse = √(P² + B²) = √(a² + b²)

then,

• secA = Hypotenuse / Base .
• cosecA = Hypotenuse / Perpendicular .

So,

• secΘ = √(a² + b²) / a
• cosecΘ = √(a² + b²) / b

Putting both values we get,

→ a{√(a² + b²) / a} - b{√(a² + b²) / b}

→ [√(a² + b²) - √(a² + b²)] (Ans.)