Prove that-itan225 - 405-tan 765- 675-0ii sin8 -3cos23 -6-cos13 -3sin35 -6-1-2iii cos 24-cos 55-cos 125-cos 204-cos 300-1-2iv tan-225- -405-tan-765- 675-0v cos 570-sin 510-sin-330- cos-390-0vi tan11 -3-2 sin4 -6-3-4cosec2-4-4 cos 217 -6-3-4 -3-2vii 3 sin-6-3-4 sin5 -5-4-1

(i)LHS =tan 225^∘ cot 405^∘ +tan 765^∘ cot 675^∘ =tan ( π+π/4 )cot ( 2 π +π/4 )+tan ( 4 π +π/4 )cot ( 4 π π/4 ) =tan π/4.cot π/4+tan π/4× ( cot π/4 ) [∵ co ...

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Byju's Answer

Standard XII

Mathematics

Trigonometric Ratios of Allied Angles

Prove that:it...

Question

Prove that:

$(i) t a n 225^{\circ} c o t 405^{\circ} + t a n 765^{\circ} c o t 675^{\circ} = 0$

$(i i) s i n \frac{8 π}{3} c o s \frac{23 π}{6} + c o s \frac{13 π}{3} s i n \frac{35 π}{6} = \frac{1}{2}$

$(i i i) c o s 24^{\circ} + c o s 55^{\circ} + c o s 125^{\circ} + c o s 204^{\circ} + c o s 300^{\circ} = \frac{1}{2}$

$(i v) t a n (- 225^{\circ}) c o t (- 405^{\circ}) - t a n (- 765^{\circ}) c o t (675^{\circ}) = 0$

$(v) c o s 570^{\circ} s i n 510^{\circ} + s i n (- 330^{\circ}) c o s (- 390^{\circ}) = 0$

$(v i) t a n \frac{11 π}{3} - 2 s i n \frac{4 π}{6} - \frac{3}{4} c o s e c^{2} \frac{π}{4} + 4 c o s^{2} \frac{17 π}{6} = \frac{3 - 4 \sqrt{3}}{2}$

$(v i i) 3 s i n \frac{π}{6} s e c \frac{π}{3} - 4 s i n \frac{5 π}{5} c o t \frac{π}{4} = 1$

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Solution

$(i) L H S = t a n 225^{\circ} c o t 405^{\circ} + t a n 765^{\circ} c o t 675^{\circ}$

$= t a n (π + \frac{π}{4}) c o t (2 π + \frac{π}{4}) + t a n (4 π + \frac{π}{4}) c o t (4 π - \frac{π}{4})$

$= t a n \frac{π}{4} . c o t \frac{π}{4} + t a n \frac{π}{4} \times (- c o t \frac{π}{4})$

$[∵ c o t (4 π - \frac{π}{4}) = - c o t \frac{π}{4}]$

=1.1 +1.(-1)

=RHS Hence proved.

$(i i) L H S = s i n \frac{8 π}{3} c o s \frac{23 π}{6} + c o s \frac{13 π}{3} s i n \frac{35 π}{6}$

$= s i n (3 π - \frac{π}{3}) c o s (4 π - \frac{π}{6}) + c o s (4 π + \frac{π}{3}) s i n (6 π - \frac{π}{6})$

$= s i n \frac{π}{3} c o s \frac{π}{6} + c o s \frac{π}{3} (- s i n \frac{π}{6})$

$[∵ s i n (6 π - θ) = - s i n θ]$

$= \frac{\sqrt{3}}{2} \times \frac{\sqrt{3}}{22} + \frac{1}{2} \times (\frac{- 1}{2})$

$= \frac{3}{4} - \frac{1}{4}$

$= \frac{2}{4}$

$= \frac{1}{2} = R H S$ Hence proved.

$(i i i) L H S = c o s 24^{\circ} + c o s 55^{\circ} + c o s 125^{\circ} + c o s 204^{\circ} + c o s 300^{\circ}$

$= c o s 24^{\circ} + c o s 204^{\circ} + c o s 55^{\circ} + c o s 125^{\circ} + c o s 300^{\circ}$

$= c o s 24^{\circ} + c o s (π + 24^{\circ}) + c o s 55^{\circ} + c o s (π - 55^{\circ}) + c o s (2 π - \frac{π}{3})$

$= c o s 24^{\circ} - c o s 24^{\circ} + c o s 55^{\circ} - c o s 55^{\circ} + c o s \frac{π}{3}$

$= c o s \frac{π}{3}$

$= \frac{1}{2} = R H S$ Hence proved.

$(i v) L H S = t a n (- 225^{\circ}) c o t (- 405^{\circ}) - t a n (- 765^{\circ}) c o t (675^{\circ})$

$= - t a n 225^{\circ} (- c o t 405^{\circ}) + t a n 765^{\circ} c o t 765^{\circ}$

$[∵ t a n (- θ) = - t a n θ$ and $c o t (- θ) = - c o t θ]$

$= t a n (π + \frac{π}{4}) c o t (2 π \frac{π}{4}) + t a n (4 π + \frac{π}{4}) c o t (4 π - \frac{π}{4})$

$= t a n \frac{π}{4} c o t \frac{π}{4} + t a n \frac{π}{4} \times (- c o t \frac{π}{4}) [∵ c o t (4 π - θ) = - c o t θ]$

=1.1+1(-1)=1-1=0=RHS Hence proved.

$(v) L H S = c o s 570^{\circ} s i n 510^{\circ} + s i n (- 330^{\circ}) c o s (- 390^{\circ})$

$= c o s (3 π + \frac{π}{6}) s i n (3 π - \frac{π}{6}) - s i n 330^{\circ} c o s 390^{\circ}$

$[∵ s i n (- θ) = - s i n θ a n d c o s (- θ) = c o s θ]$

$= - c o s \frac{π}{6} s i n \frac{π}{6} - s i n (2 π - \frac{π}{6}) c o s (2 π + \frac{π}{6})$

$= - s i n \frac{π}{6} c o s \frac{π}{6} + s i n \frac{π}{6} c o s \frac{π}{6} [∵ s i n (2 π - θ) = - s i n θ]$

0=RHS Hence proved.

$(v i) L H S = t a n \frac{11 π}{3} - 2 s i n \frac{4 π}{6} - \frac{3}{4} c o s e c^{2} \frac{π}{4} + 4 c o s^{2} \frac{17 π}{6}$

$= t a n (4 π - \frac{π}{3}) - 2 s i n \frac{2 π}{3} - \frac{3}{4} \times (\sqrt{2})^{2} + 4 c o s^{2} (3 π - \frac{π}{6})$

$= - t a n \frac{π}{3} - 2 s i n (π - \frac{π}{3}) - \frac{3}{4} \times 2 + 4 c o s^{2} \frac{π}{6}$

$(∵ t a n (4 π - \frac{π}{3}) = - t a n \frac{π}{3}, c o s (3 π - \frac{π}{6}) = - c o s \frac{π}{6})$

$= - \sqrt{3} - 2 s i n \frac{π}{3} - \frac{3}{2} + 4 \times {(\frac{\sqrt{3}}{2})}^{2}$

$= - \sqrt{3} - 2 \times \frac{\sqrt{3}}{2} - \frac{3}{2} + 4 \times \frac{3}{4}$

$= - \sqrt{3} - \sqrt{3} - \frac{3}{2} + 3 = - 2 \sqrt{3} \frac{- 3 + 6}{2} = - 2 \sqrt{3} \frac{3}{2}$

$= \frac{3 - 4 \sqrt{3}}{2}$ =RHS Hence proved.

$(v i i) L H S = 3 s i n \frac{π}{6} s e c \frac{π}{3} - 4 s i n \frac{5 π}{6} c o t \frac{π}{4}$

$= 3 \times \frac{1}{2} \times 2 - 4 s i n (π - \frac{π}{6}) \times 1$

$= 3 - 4 s i i n \frac{π}{6} [∵ s i n (π - θ) = s i n θ]$

$= 3 - 4 \times \frac{1}{2} = 3 - 2 = 1$ =RHS Hence proved.

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