Determine the smallest positive value of x in degrees for which tanx-100-tanx-50- tan x tanx-50-Prove that cos 7 x-cos 8 x-1-2 cos 5 x-cos 2 x-cos 3 x

We have, tan (x+100^∘) = tan (x+50^∘)tan x tan (x 50^∘) ⇒ tan (x+100^∘)/tan (x 50^∘)=tan (x+50^∘)tan x ⇒ sin (x+100^∘)cos (x 50^∘)/cos(x+100^∘)sin (x 50^∘)=s ...

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Byju's Answer

Standard XI

Mathematics

Trigonometric Ratios Using Right Angled Triangle

Determine the...

Question

Determine the smallest positive value of x (in degrees) for which tan $(x + 100^{\circ}) = t a n (x + 50^{\circ}) t a n x t a n (x - 50^{\circ})$ .

Prove that $\frac{c o s 7 x - c o s 8 x}{1 + 2 c o s 5 x} = c o s 2 x - c o s 3 x$

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Solution

We have,

$t a n (x + 100^{\circ}) = t a n (x + 50^{\circ}) t a n x t a n (x - 50^{\circ})$

$\Rightarrow \frac{t a n (x + 100^{\circ})}{t a n (x - 50^{\circ})} = t a n (x + 50^{\circ}) t a n x$

$\Rightarrow \frac{s i n (x + 100^{\circ}) c o s (x - 50^{\circ})}{c o s (x + 100^{\circ}) s i n (x - 50^{\circ})} = \frac{s i n (x + 50^{\circ}) s i n x}{c o s (x + 50^{\circ}) c o s x}$

Applying componendo and dividendo, we get

$\frac{s i n (x + 100^{\circ}) c o s (x - 50^{\circ}) + c o s (x + 100^{\circ}) s i n (x - 50^{\circ})}{s i n (x + 100^{\circ}) c o s (x - 50^{\circ}) - c o s (x + 100^{\circ}) s i n (x - 50^{\circ})}$

$= \frac{s i n (x + 50^{\circ}) s i n x + c o s (x + 50^{\circ}) c o s x}{s i n (x + 50^{\circ}) s i n x - c o s (x + 50^{\circ}) c o s x}$

$\Rightarrow \frac{s i n (x + 100^{\circ} + x - 50^{\circ})}{s i n (x + 100^{\circ} - x + 50^{\circ})} = \frac{c o s (x + 50^{\circ} - x)}{- c o s (x + 50^{\circ} + x)}$

$∵ s i n (A \pm B) = s i n A c o s B \pm c o s A s i n B$
$and c o s (A \pm B) = c o s A c o s B \mp s i n A s i n B$

$\Rightarrow \frac{s s i n (2 x + 50^{\circ})}{s i n 150^{\circ}} = \frac{c o s 50^{\circ}}{- c o s (2 x + 50^{\circ})}$

$\Rightarrow s i n (2 x + 50^{\circ}) c o s (2 x + 50^{\circ}) = - s i n 150^{\circ} c o s 50^{\circ}$

$\Rightarrow 2 s i n (2 x + 50^{\circ}) c o s (2 x + 50^{\circ}) = - 2 \times \frac{1}{2} c o s 50^{\circ}$

$[∵ s i n 150^{\circ} = \frac{1}{2}]$

$\Rightarrow s i n (4 x + 100^{\circ}) = - c o s 50^{\circ}$

$[∵ 2 s i n A c o s A = s i n 2 A]$

$\Rightarrow s i n (4 x + 100^{\circ}) = s i n (270^{\circ} - 50^{\circ})$

$\Rightarrow 4 x + 100^{\circ} = 270^{\circ} - 50^{\circ}$

$\Rightarrow 4 x = 270^{\circ} - 50^{\circ} - 100^{\circ}, 4 x = 270^{\circ} - 150^{\circ}$

$\Rightarrow 4 x = 120^{\circ}$

$\Rightarrow x = 30^{\circ}$
II part-

We have, $L H S = \frac{c o s 7 x - c o s 8 x}{1 + 2 c o s 5 x}$

Multiply numerator and denominator by 2 $s i n \frac{5 x}{2}$ , we get

$L H S = \frac{2 s i n \frac{5 x}{2} c o s 7 x - 2 s i n \frac{5 x}{2} c o s 8 x}{2 s i n \frac{5 x}{2} + 4 s i n \frac{5 x}{2} c o s 5 x}$

$= \frac{(s i n \frac{19 x}{2} - s i n \frac{9 x}{2}) - (s i n \frac{21 x}{2} - s i n \frac{11 x}{2})}{2 s i n \frac{5 x}{2} + 2 s i n \frac{15 x}{2} - 2 s i n \frac{5 x}{2}}$

[ $∵$ 2 sin A cos B = sin (A + B) + sin (A - B)]

$= \frac{(s i n \frac{19 x}{2} + s i n \frac{11 x}{2}) - (s i n \frac{9 x}{2} + s i n \frac{21 x}{2})}{2 s s i n \frac{15 x}{2}}$

$= \frac{2 s i n \frac{15 x}{2} c o s 2 x - 2 s i n \frac{15 x}{2} c o s 3 x}{2 s i n \frac{15 x}{2}}$

$[∵ s i n C + s i n D = 2 s i n \frac{C + D}{2} c o s \frac{C - D}{2}]$

$= cos 2 x - cos 3 x = R H S$ Hence proved.

Suggest Corrections

Determine the smallest positive value of x (in degrees) for which tan (x+100∘)=tan(x+50∘)tan x tan (x−50∘). Prove that cos 7x−cos 8x1+2 cos 5x=cos 2x−cos 3x

Determine the smallest positive value of x (in degrees) for which tan $(x + 100^{\circ}) = t a n (x + 50^{\circ}) t a n x t a n (x - 50^{\circ})$ .

Prove that $\frac{c o s 7 x - c o s 8 x}{1 + 2 c o s 5 x} = c o s 2 x - c o s 3 x$