Question

show that, 1-sin 60°/cos 60°=tan 60°-1/tan 60°+1​

Answer 1

Answer:

Step-by-step explanation:

1-sin60°/cos60°

=(1-√3/2)/(1/2)

=(2-√3)/2×2/1

=2-√3

tan60°-1/tan60°+1

=(√3-1)/(√3+1)

=(√3-1)(√3-1)/(√3+1)(√3-1)

=(3-2√3+1)/(3-1)

=(4-2√3)/2

=2(2-√3)/2

=2-√3

∴, LHS=RHS(Proved)

Answer 2

» Question :

Proof that :

$\green{\sf{\dfrac{1 - sin 60^{\circ}}{cos 60^{\circ}} = \dfrac{tan 60^{\circ} - 1}{tan 60^{\circ} + 1}}}$

» To Find :

To prove that LHS = RHS .

» We Know :

• $\sf{Cos 60^{\circ} = \dfrac{1}{2}}$

• $\sf{Sin 60^{\circ} = \dfrac{\sqrt{3}}{2}}$

• $\sf{tan 60^{\circ} = \sqrt{3}}$

» Concept :

According to the question , we have to prove that LHS is equal to RHS , so by putting the values of 60° in the Equation , we can prove that RHS = RHS.

» Solution :

Given Equation ;

$\sf{\dfrac{1 - sin 60^{\circ}}{cos 60^{\circ}} = \dfrac{tan 60^{\circ} - 1}{tan 60^{\circ} + 1}}$

Substituting the values of 60° in the Equation , we get :

$\sf{\Rightarrow \dfrac{1 - \dfrac{\sqrt{3}}{2}}{\dfrac{1}{2}} = \dfrac{\sqrt{3} - 1}{\sqrt{3} + 1}}$

$\sf{\Rightarrow \dfrac{\dfrac{2 - \sqrt{3}}{2}}{\dfrac{1}{2}} = \dfrac{\sqrt{3} - 1}{\sqrt{3} + 1}}$

$\sf{\Rightarrow \dfrac{2 - \sqrt{3}}{2} \times 2 = \dfrac{\sqrt{3} - 1}{\sqrt{3} + 1}}$

$\sf{\Rightarrow \dfrac{2 - \sqrt{3}}{\not{2}} \times \not{2} = \dfrac{\sqrt{3} - 1}{\sqrt{3} + 1}}$

$\sf{\Rightarrow 2 - \sqrt{3} = \dfrac{\sqrt{3} - 1}{\sqrt{3} + 1}}$

Multiplying (√3 - 1) on both the sides we get :

$\sf{\Rightarrow 2 - \sqrt{3} \times \left(\sqrt{3} + 1\right) = \dfrac{\sqrt{3} - 1}{\cancel{\sqrt{3} + 1}} \times \cancel{\left(\sqrt{3} + 1\right)}}$

$\sf{\Rightarrow 2\sqrt{3} - \sqrt{3} + 2 - 3 = \sqrt{3} - 1}$

$\sf{\Rightarrow 2\sqrt{3} - \sqrt{3} - 1 = \sqrt{3} - 1}$

$\sf{\Rightarrow 2\sqrt{3} = \sqrt{3} - 1 + \sqrt{3} + 1}$

$\sf{\Rightarrow 2\sqrt{3} = 2\sqrt{3} - \not{1} + \not{1}}$

$\purple{\sf{\Rightarrow 2\sqrt{3} = 2\sqrt{3}}}$

Hence , RHS = RHS .

Thus ,

$\green{\sf{\underline{\boxed{\dfrac{1 - sin 60^{\circ}}{cos 60^{\circ}} = \dfrac{tan 60^{\circ} - 1}{tan 60^{\circ} + 1}}}}}$

$\boxed{Proved}$

» Additional information :

• sin(a + b) = sinAcosB + cosAsinB

• sin(a - b) = sinAcosB - cosAsinB

• cos(a + b) = cosAcosB - sinAsinB

• cos(a - b) = cosAcosB + sinAsinB

• sin2A = 2sinAcosA