Prove that -3 cosec 20- sec 20-4-OrIf cos -4-5- sin -5-13 and - - lie between 0 and -4- find tan 2 -

LHS= √(3) cosec 20^∘ sec 20^∘=tan 60^∘ cosec 20^∘ sec 20^∘ = sin 60^∘/cos 60^∘1/sin 20^∘ 1/cos 20^∘ = sin 60^∘ cos 20^∘ cos 60^∘ sin 20^∘/cos 60^∘ sin20^∘ c ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XI

Mathematics

Trigonometric Ratios of Compound Angles

Prove that √3...

Question

Prove that $\sqrt{3} c o s e c 20^{\circ} s e c 20^{\circ} = 4$ .

Or

If $c o s (α + β) = \frac{4}{5}, s i n (α - β) = \frac{5}{13}$ and $α, β$ lie between 0 and $\frac{π}{4}$ , find $t a n 2 α$ .

Open in App

Solution

LHS= $\sqrt{3} c o s e c 20^{\circ} - s e c 20^{\circ} = t a n 60^{\circ} c o s e c 20^{\circ} - s e c 20^{\circ}$

$= \frac{s i n 60^{\circ}}{c o s 60^{\circ}} \frac{1}{s i n 20^{\circ}} - \frac{1}{c o s 20^{\circ}}$

$= \frac{s i n 60^{\circ} c o s 20^{\circ} - c o s 60^{\circ} s i n 20^{\circ}}{c o s 60^{\circ} s i n 20^{\circ} c o s 20^{\circ}} = \frac{s i n (60^{\circ} - 20^{\circ})}{c o s 60^{\circ} s i n 20^{\circ} c o s 20^{\circ}}$

$= \frac{s i n 40^{\circ}}{c o s 60^{\circ} s i n 20^{\circ} c o s 20^{\circ}}$

$= \frac{2 s i n 20^{\circ} c o s 20^{\circ}}{\frac{1}{2} \times s i n 20^{\circ} c o s 20^{\circ}}$

=4=RHS

We have $c o s (α + β) = \frac{4}{5}$ and $s i n (α - β) = \frac{5}{13}$

$∴ t a n (α + β) = \frac{s i n (α + β)}{c o s (α + β)} = \frac{\sqrt{1 - c o s^{2} (α + β)}}{c o s (α + β)} = \frac{\sqrt{1 - {(\frac{4}{5})}^{2}}}{\frac{4}{5}} = \frac{\sqrt{\frac{9}{25}}}{\frac{4}{5}} = \frac{3}{4}$

and $t a n (α - β) = \frac{s i n (α - β)}{\sqrt{1 - s i n^{2} (α - β)}} = \frac{\frac{5}{13}}{\sqrt{1 - {(\frac{5}{13})}^{2}}} = \frac{\frac{5}{13}}{\sqrt{\frac{169 - 25}{169}}} = \frac{\frac{5}{13}}{\frac{12}{13}} = \frac{5}{12}$

Now, $t a n 2 α = t a n (α + β + α - β)$

$= \frac{t a n (α + β) + t a n (α - β)}{1 - t a n (α + β) . t a n (α - β)} = \frac{\frac{3}{4} + \frac{5}{12}}{1 - \frac{3}{4} \times \frac{5}{12}} = \frac{36 + 20}{48 - 15}$

Hence, $t a n 2 α = \frac{56}{33}$

Suggest Corrections

Similar questions

Q. If

cos (α + β) = 4 / 5, sin (α - β) = 5 / 13

and

α, β

lie between

0

and

\frac{π}{4}

, find

tan 2 α

Q. If

cos (α + β) = \frac{4}{5}, sin (α - β) = \frac{5}{13}

and

α, β

lie between 0 and

\frac{π}{4}

, then

tan 2 α =

Q. If

0 < α, β < \frac{π}{4}, cos (α + β) = \frac{4}{5}, sin (α - β) = \frac{5}{13}

, then

tan 2 α =

Q. If

cos (α + β) = \frac{4}{5}, sin (α - β) = \frac{5}{13} a n d α, β

lie between

0

and

\frac{π}{4}

; then find

tan 2 α

Q. Let

cos (α + β) = \frac{4}{5}

and let

sin (α - β) = \frac{5}{13}

, where

0 \leq α, β \leq \frac{π}{4}

. Then

tan 2 α =

Join BYJU'S Learning Program

Prove that √3 cosec 20∘sec 20∘=4. Or If cos (α+β)=45, sin(α−β)=513 and α, β lie between 0 and π4, find tan 2α.

Prove that $\sqrt{3} c o s e c 20^{\circ} s e c 20^{\circ} = 4$ .

Or

If $c o s (α + β) = \frac{4}{5}, s i n (α - β) = \frac{5}{13}$ and $α, β$ lie between 0 and $\frac{π}{4}$ , find $t a n 2 α$ .