Question

If bcos A = a, then prove that cosec A + cot A =
$\sqrt{ \frac{b + a}{b - a} }$
​

Answer 1

Answer:

Given

bcos A = a

\underline{\underline{\bf{\maltese\:\:To\: Prove}}}

✠ToProve

cosec A + cot A = \sqrt{ \frac{b + a}{b - a} }

b−a

b+a

\underline{\underline{\bf{\maltese\:Formula \: used}}}

✠Formulaused

AC² = AB² - BC²

\underline{\underline{\bf{\maltese\:Proof}}}

✠Proof

⇢bcos A = a

⇢cos A = \bf \frac{a}{b}

b

a

AC² = AB² - BC²

AC² = b² - a²

AC = \bf\sqrt{ {b}^{2} - {a}^{2} }

b

2

−a

2

⇢cosec A = \bf\frac{hypotenuse}{perpendicular}

perpendicular

hypotenuse

⇢cosec A = \bf\frac{b}{ \sqrt{ {b}^{2} - {a}^{2} } }

b

2

−a

2

b

⇢cot A = \bf\frac{base}{perpendicular}

perpendicular

base

⇢cot A = \bf\frac{a}{ \sqrt{ {b}^{2} - {a}^{2} } }

b

2

−a

2

a

⇢cosec A + cot A = \bf\frac{b + a}{ \sqrt{ {b}^{2} - {a}^{2} } }

b

2

−a

2

b+a

⇢cosec A + cot A = \bf\sqrt{ \frac{b \: + a}{b - a} }

b−a

b+a

HENCE PROVED !!