Question

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Answer 1

☆ Question (1) :

If $\sf{tan35^{\circ} = a}$ , then Prove that

$\sf{\dfrac{tan145^{\circ} - tan125^{\circ}}{1 + tan145^{\circ}tan125^{\circ}} = \dfrac{1 - a^{2}}{2a}}$

☆ To Find :

To Proof that LHS = RHS.

☆ We Know :

• $\sf{\dfrac{tanA - tanB}{1 + tanAtanB} = tan(A - B)}$

• $\sf{\dfrac{2tanA}{1 - tan^{2}A} = tan2A}$

• $\sf{tanAtan(90^{\circ} - A) = 1}$

☆ Concept :

We Know that ,

$\sf{\dfrac{2tanA}{1 - tan^{2}A} = tan2A}$

so, by dividing 1 by both the sides, we get :

$\implies \sf{\dfrac{1}{\dfrac{2tanA}{1 - tan^{2}A}} = \dfrac{1}{tan2A}} \\ \\ \\ \\ \implies \sf{\dfrac{1 - tan^{2}A}{2tanA} = \dfrac{1}{tan2A}} \\ \\ \\ \\ \therefore \purple{\sf{\dfrac{1}{tan2A} = \dfrac{1 - tan^{2}A}{2tanA}}}$

Now, by simplifying the Equation in the Formulae we can Proof that LHS = RHS.

☆ Solution :

Given :

• $\sf{tan35^{\circ} = a}$

Calculation :

$\sf{\dfrac{tan145^{\circ} - tan125^{\circ}}{1 + tan145^{\circ}tan125^{\circ}} = \dfrac{1 - a^{2}}{2a}}$

Using the formula ,

$\sf{\dfrac{tanA - tanB}{1 + tanAtanB} = tan(A - B)}$

We get :

$\implies \sf{tan(145^{\circ} - 125^{\circ}) = \dfrac{1 - a^{2}}{2a}} \\ \\ \\ \\ \implies \sf{tan20^{\circ} = \dfrac{1 - a^{2}}{2a}}$

Putting the value of a in the Equation , we get :

$\implies \sf{tan20^{\circ} = \dfrac{1 - tan^{2}35^{\circ}}{2tan35^{\circ}}} \\ \\ \\ \implies \sf{tan20^{\circ} = \dfrac{1 - tan(35^{\circ})^{2}}{2tan35^{\circ}}} \\ \\ \\ \sf{tan20^{\circ} = \dfrac{1 - tan^{2}35^{\circ}}{2tan35^{\circ}}}$

Using the formula ,

$\sf{\dfrac{1}{tan2A} = \dfrac{1 - tan^{2}A}{2tanA}}$

We get :

$\implies \sf{tan20^{\circ} = \dfrac{1}{tan2 \times 35^{\circ}}} \\ \\ \\ \\ \implies \sf{tan20^{\circ} = \dfrac{1}{tan70^{\circ}}}$

By multiplying 70° on both the sides, we get :

$\implies \sf{tan20^{\circ} \times 70^{\circ} = \dfrac{1}{tan70^{\circ}} \times 70^{\circ}} \\ \\ \\ \implies \sf{tan20^{\circ}tan70^{\circ} = 1}$

Using the formula ,

$\sf{tanAtan(90^{\circ} - A) = 1}$

We get ,

$\implies \sf{tan20^{\circ}tan(90^{\circ} - 20^{\circ} = 1} \\ \\ \\ \\ \implies \sf{1 = 1} \\ \\ \\ \therefore \purple{\sf{1 = 1}}$

Hence , LHS = RHS is proved.

☆ Question (2) :

Evaluate : $\sf{cos75^{\circ}}$

☆ To Find :

The value of $\sf{cos75^{\circ}}$.

☆ We Know :

• $\sf{cos(A + B) = cosAcosB - sinAsinB}$

☆ Concept :

According to the question , we have to find the value of $\sf{cos75^{\circ}}$ , and we know that it can be also written as $\sf{cos(45^{\circ} + 30^{\circ})}$

So by solving this , we can find the value of $\sf{cos75^{\circ}}$

☆ Calculation :

We Know :

• $\sf{cos45^{\circ} = \dfrac{1}{\sqrt{2}}}$

• $\sf{cos30^{\circ} = \dfrac{\sqrt{3}}{2}}$

• $\sf{sin30^{\circ} = \dfrac{1}{\sqrt{2}}}$

• $\sf{sin45^{\circ} = \dfrac{1}{\sqrt{2}}}$

Equation :

$\purple{\sf{cos(45^{\circ} + 30^{\circ})}}$

By using the formula ,

$\sf{cos(A + B) = cosAcosB - sinAsinB}$

We get ,

$\implies \sf{cos45^{\circ}cos30^{\circ} - sin45^{\circ}sin30^{\circ}}$

Putting the value of cos45° and 30° in the Equation , we get :

$\implies \sf{\dfrac{1}{\sqrt{2}} \times \dfrac{\sqrt{3}}{2} - \dfrac{1}{\sqrt{2}} \times \dfrac{1}{2}} \\ \\ \\ \implies \sf{\dfrac{\sqrt{3}}{2\sqrt{2}} - \dfrac{1}{2\sqrt{2}}} \\ \\ \\ \implies \sf{\dfrac{\sqrt{3} - 1}{2\sqrt{2}}} \\ \\ \\ \therefore \purple{\sf{cos75^{\circ} = \dfrac{\sqrt{3} - 1}{2\sqrt{2}}}}$

Hence , the value of $\sf{cos75^{\circ}}$ is $\dfrac{\sqrt{3} - 1}{2\sqrt{2}}$

Answer 2

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1)

GIVEN:-

$\bf{ \tan35 \degree = a }$

PROVE THAT:-

$\bf{ \frac{ \tan45 \degree - \tan125 \degree }{1 + \tan145 \degree \tan125 \degree } } = \frac{1 - {a}^{2} }{2a}$

SOLUTION:-

$\bf{ \tan145 \degree = \tan(180 \degree - 35 \degree) = - \tan35 \degree = - a, }$

$\bf{ \tan125 \degree = \tan(90 \degree + 35\degree) = - \cot35 \degree = - \frac{1}{ \tan35 \degree} = - \frac{1}{a} }$

$\bf{ L.H.S = \frac{ \tan145 \degree - \tan125 \degree}{1 + \tan 145 \degree \tan125 \degree} = \frac{( - a) - ( \frac{ - 1}{a}) }{1 + ( - a)( \frac{ - 1}{a} )} = \frac{ - a + \frac{1}{a} }{1 + 1} }$

$\bf{ = \frac{ { - a}^{2} + 1 }{2a} = \frac{1 - {a}^{2} }{2a} = R.H.S }$

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2)

$\bf{ \cos75 \degree}$

$\bf{ = \cos(30 + 45)}$

= cos 30 cos 45 - sin 30 sin 45

( °.° cos (A+B) = cos A cos B - sin A sin B)

$\bf{\frac{√3}{2}.\frac{1}{√2}-\frac{1}{2}.\frac{1}{√2}}$

= $\bf{\frac{√3-1}{2√2}}$

$\bf{\frac{√3-1}{2√2}\times\frac{√2}{√2}}$

$\bf{\frac{√6-√2}{4}}$

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