Question

Probe that (1+cotx-cosecx)(1+tanx+secx)=2

Answer 1

$\huge\mathfrak{Answer:}$

Given:

• We have been given some trigonometric ratios: (1 + cotx - cosecx ) (1 + tanx + secx ).

To Prove:

• We need to prove that (1 + cotx - cosecx ) (1 + tanx + secx ) = 2.

Proof:

The given trignometric ratio is

(1 + cotx - cosecx ) (1 + tanx + secx ).

Taking LHS, We have

$\sf{(1 + \cot(x) - \cos(x) )(1 + \tan(x) + \sec(x) )}$

$\sf{(1 + \dfrac{ \cos(x) }{ \sin(x) } - \dfrac{1}{ \sin(x) } ) (1 + \dfrac{ \sin(x) }{ \cos(x) } + \frac{1}{ \cos(x) } ) }$

$\sf{ (\dfrac{ \sin(x )+ \cos(x) - 1 }{ \sin(x) })}( \dfrac{ \cos(x) + \sin(x) + 1}{ \cos(x) }$

$\sf \dfrac{(sin\ x\ +\ cos\ x)^2\ -\ 1}{sinx \times cosx}$

$\sf{\dfrac{\sin ^2 x \: + \: \cos ^2 x \: + \: 2 \: \sin x \cos x \: - \: 1}{\sin x \cos x}}$

$\sf{\dfrac{1 \: + \: 2 \cancel{\sin x \cos x} \: - \: 1}{\cancel{\sin x \cos x}}}$

$\sf{2 = 2}$

LHS = RHS

Hence proved!!

Answer 2

★ Question ★

→ (1+cotx-cosecx)(1+tanx+secx)=2

◇ To prove

→ (1+cotx-cosecx)(1+tanx+secx)=2

Solution ↑ (refer to the attachment.)