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Question Number 123232 by mohammad17 last updated on 24/Nov/20

$$provethatcot5x=\frac{1-10{tan}^{2}x+5{tan}^{4}x}{{tan}^{5}x-10{tan}^{3}x+5tanx}?$$

Answered by $@y@m last updated on 24/Nov/20

$$\tan (2x+3x)=\frac{\tan 2x+\tan 3x}{1-\tan 2x.\tan 3x}$$$$\Rightarrow \tan 5x=\frac{\frac{2\tan x}{1-\tan {}^{2}x}+\frac{3\tan x-\tan {}^{3}x}{1-3\tan {}^{2}x}}{1-\left(\frac{2\tan x}{1-\tan {}^{2}x}\right)\left(\frac{3\tan x-\tan {}^{3}x}{1-3\tan {}^{2}x}\right)}$$$$=\frac{2\tan x(1-3\tan {}^{2}x)+(1-\tan {}^{2}x)(3\tan x-\tan {}^{3}x)}{(1-\tan {}^{2}x)(1-3\tan {}^{2}x)-2\tan x(3\tan x-\tan {}^{3}x)}$$$$=\frac{2\tan x-{6\tan }^{3}x+3\tan x-{3\tan }^{3}x-{\tan }^{3}x+{\tan }^{5}x}{1-\tan {}^{2}x-3\tan {}^{2}x+3\tan {}^{4}x-{6\tan }^{2}x+{2\tan }^{4}x}$$$$=\frac{\tan {}^{5}x-10\tan {}^{3}x+5\tan x}{1-10\tan {}^{2}x+5\tan {}^{4}x}$$

Commented by mohammad17 last updated on 24/Nov/20

$$thankyousir$$

Answered by som(math1967) last updated on 24/Nov/20

$$\cot 5\mathrm{x}$$$$=\frac{1}{\tan (3\mathrm{x}+2\mathrm{x})}=\frac{1-\tan 3\mathrm{xtan}2\mathrm{x}}{\tan 3\mathrm{x}+\tan 2\mathrm{x}}$$$$=\frac{1-\frac{(3\mathrm{tanx}-{\tan }^3\mathrm{x})\left(2\mathrm{tanx}\right)}{(1-{3\tan }^2\mathrm{x})(1-{\tan }^2\mathrm{x})}}{\frac{3\mathrm{tanx}-{\tan }^3\mathrm{x}}{1-{3\tan }^2\mathrm{x}}+\frac{2\mathrm{tanx}}{1-{\tan }^2\mathrm{x}}}$$$$\mathrm{let}\mathrm{tanx}=\mathrm{a}$$$$\frac{(1-{3\mathrm{a}}^2)(1-{\mathrm{a}}^2)-{6\mathrm{a}}^2+{2\mathrm{a}}^4}{(1-{\mathrm{a}}^2)(3\mathrm{a}-{\mathrm{a}}^3)+2\mathrm{a}-{6\mathrm{a}}^3}$$$$=\frac{1-{\mathrm{a}}^2-{3\mathrm{a}}^2+{3\mathrm{a}}^4-{6\mathrm{a}}^2+{2\mathrm{a}}^4}{3\mathrm{a}-{\mathrm{a}}^3-{3\mathrm{a}}^3+{\mathrm{a}}^5+2\mathrm{a}-{6\mathrm{a}}^3}$$$$=\frac{1-{10\mathrm{a}}^2+{5\mathrm{a}}^4}{5\mathrm{a}-{10\mathrm{a}}^3+{\mathrm{a}}^5}$$$$\mathrm{now}\mathrm{put}\mathrm{a}=\mathrm{tanx}$$

Commented by mohammad17 last updated on 24/Nov/20

$$thankyousir$$

Answered by TANMAY PANACEA last updated on 24/Nov/20

$$tan\left(nx\right)=\frac{s_1-s_3+s_5-s_7+...}{1-s_2+s_4-s_6+..}$$$$tan5x=\frac{s_1-s_3+s_5}{1-s_2+s_4}$$$$s_1=5C_{1}tanx=5tanx$$$$s_3=5C_3{tan}^{3}x=\frac{5\times 4}{2}{tan}^{3}x=10{tan}^{3}x$$$$s_5=5C_5{tan}^{5}x={tan}^{5}x$$$$s_2=5c_2{tan}^{2}x=10{tan}^{2}x$$$$s_4=5c_4{tan}^{4}x=5{tan}^{4}x$$$$cot5x=\frac{1}{tan5x}=\frac{1-10{tan}^{2}x+5{tan}^{4}x}{5tanx-10{tan}^{3}x+{tan}^{5}x}$$

Commented by som(math1967) last updated on 25/Nov/20

$$\mathrm{Nice}\mathrm{sir}$$

Commented by som(math1967) last updated on 25/Nov/20

খুব ভাল

Commented by TANMAY PANACEA last updated on 25/Nov/20

$$thankyou...$$

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