Question

$\huge{Question¶}$​

Answer 1

Answer:

${\boxed{\boxed{\begin{array}{cc}\bf \: \to \:given \: that : \\ \\ \alpha + \beta \neq \: 0 \\ \\ \sf \: also \: given : \\ \\ \rm \: sin \: \alpha + sin \: \beta = 2 \: sin( \alpha + \beta ) \\ \\ \red{ \underline{ \bf \: we \: have \: to \: prove : }} \\ \\ \rm \: tan \: \frac{ \alpha }{2}.tan \: \frac{ \beta }{2} = \frac{1}{3} \end{array}}}}$

${\boxed{\boxed{\begin{array}{cc} \red { \underline{\bf \: solution}} \\ \\ \rm \: sin \: \alpha + \: sin \: \beta = 2 \: sin( \alpha + \beta ) \\ \\ \orange{{\boxed{\begin{array}{cc}\bf \: we \: know : \\ \\ \to \: \bf \: sin \: x + sin \: y = 2 \: sin \: ( \frac{x + y}{2}) . \: cos \: ( \frac{x - y}{2} ) \\ \\ \rm \: \to \: sin \: 2x = 2 \: sin \: x \: cos \: x \\ \therefore \: \rm \: sin \: x = 2 \: sin \: \frac{x}{2} . \: cos \: \frac{x}{2} \end{array}}}} \\ \pink{ \sf \: apply \: this} \\ \\ \rm \implies \: \cancel2 \: sin \: ( \frac{ \alpha + \beta }{2} ). cos \: ( \frac{ \alpha - \beta }{2}) = \cancel 2 \times 2 \: sin ( \frac{ \alpha + \beta }{2} ).cos( \frac{ \alpha + \beta }{2} ) \\ \\ \rm \implies \:sin( \frac{ \alpha + \beta }{2} ) \{ cos( \frac{ \alpha - \beta }{2} )- 2 \: cos( \frac{ \alpha + \beta }{2} ) \} = 0 \\ \end{array}}}}$

${\boxed{\boxed{\begin{array}{cc}\sf \: since \: \: \alpha + \beta \neq \: 0 \: \: \: sin( \frac{ \alpha + \beta }{2}) \neq0 \\ \\ \\ \therefore \: \rm \: cos( \frac{ \alpha - \beta }{2} ) - 2 \: cos( \frac{ \alpha + \beta }{2} ) = 0 \\ \\ \orange{{\boxed{\begin{array}{cc}\bf \: we \: know: \\ \\ \rm \: \to \: cos(x \pm y) = cos \: x \: cos \: y \mp \: sin \: x \: sin \: y\end{array}}}} \\ \pink{ \sf \: apply \: this} \\ \\ \rm \implies \: cos \frac{ \alpha }{2} .cos \frac{ \beta }{2} + sin \frac{ \alpha }{2} .sin \frac{ \beta }{2} - 2(cos \frac{ \alpha }{2} .cos \frac{ \beta }{2} - sin \frac{ \alpha }{2}.sin \frac{ \beta }{2} ) = 0 \\ \\ \rm \implies \: 3 \: sin \frac{ \alpha }{2} \: sin \frac{ \beta }{2} = cos \frac{ \alpha }{2} .cos \frac{ \beta }{2} \\ \\ \rm \implies \: \frac{sin \frac{ \alpha }{2} .sin \frac{ \beta }{2} }{cos \frac{ \alpha }{2}.cos \frac{ \beta }{2} } = \frac{1}{3} \\ \\ \rm \implies \: tan \: \frac{ \alpha }{2} .tan \frac{ \beta }{2} = \frac{1}{3} \\ \\ \therefore \: L.H.S = R.H.S\end{array}}}}$

Answer 2

Explanation:

Step-by-step explanation:

Hey mate

Here is the answer

EXPLANATION.

${\displaystyle {\begin{aligned}\tan x&{}=\sum _{n=0}^{\infty }{\frac {U_{2n+1}}{(2n+1)!}}x^{2n+1}\\[8mu]&{}=\sum _{n=1}^{\infty }{\frac {(-1)^{n-1}2^{2n}\left(2^{2n}-1\right)B_{2n}}{(2n)!}}x^{2n-1}\\[5mu]&{}=x+{\frac {1}{3}}x^{3}+{\frac {2}{15}}x^{5}+{\frac {17}{315}}x^{7}+\cdots ,\qquad {\text{for }}|x|<{\frac {\pi }{2}}.\end{aligned}}}{\displaystyle {\begin{aligned}\tan x&{}=\sum _{n=0}^{\infty }{\frac {U_{2n+1}}{(2n+1)!}}x^{2n+1}\\[8mu]&{}=\sum _{n=1}^{\infty }{\frac {(-1)^{n-1}2^{2n}\left(2^{2n}-1\right)B_{2n}}{(2n)!}}x^{2n-1}\\[5mu]&{}=x+{\frac {1}{3}}x^{3}+{\frac {2}{15}}x^{5}+{\frac {17}{315}}x^{7}+\cdots ,\qquad {\text{for }}|x|<{\frac {\pi }{2}}.\end{aligned}}}$

${\displaystyle {\begin{aligned}\csc x&=\sum _{n=0}^{\infty }{\frac {(-1)^{n+1}2\left(2^{2n-1}-1\right)B_{2n}}{(2n)!}}x^{2n-1}\\[5mu]&=x^{-1}+{\frac {1}{6}}x+{\frac {7}{360}}x^{3}+{\frac {31}{15120}}x^{5}+\cdots ,\qquad {\text{for }}0<|x|<\pi .\end{aligned}}}$

${\displaystyle {\begin{aligned}\csc x&=\sum _{n=0}^{\infty }{\frac {(-1)^{n+1}2\left(2^{2n-1}-1\right)B_{2n}}{(2n)!}}x^{2n-1}\\[5mu]&=x^{-1}+{\frac {1}{6}}x+{\frac {7}{360}}x^{3}+{\frac {31}{15120}}x^{5}+\cdots ,\qquad {\text{for }}0<|x|<\pi .\end{aligned}}}$

${\displaystyle {\begin{aligned}\sec x&=\sum _{n=0}^{\infty }{\frac {U_{2n}}{(2n)!}}x^{2n}=\sum _{n=0}^{\infty }{\frac {(-1)^{n}E_{2n}}{(2n)!}}x^{2n}\\[5mu]&=1+{\frac {1}{2}}x^{2}+{\frac {5}{24}}x^{4}+{\frac {61}{720}}x^{6}+\cdots ,\qquad {\text{for }}|x|<{\frac {\pi }{2}}.\end{aligned}}}$

${\displaystyle {\begin{aligned}\sec x&=\sum _{n=0}^{\infty }{\frac {U_{2n}}{(2n)!}}x^{2n}=\sum _{n=0}^{\infty }{\frac {(-1)^{n}E_{2n}}{(2n)!}}x^{2n}\\[5mu]&=1+{\frac {1}{2}}x^{2}+{\frac {5}{24}}x^{4}+{\frac {61}{720}}x^{6}+\cdots ,\qquad {\text{for }}|x|<{\frac {\pi }{2}}.\end{aligned}}}$

${\displaystyle {\begin{aligned}\cot x&=\sum _{n=0}^{\infty }{\frac {(-1)^{n}2^{2n}B_{2n}}{(2n)!}}x^{2n-1}\\[5mu]&=x^{-1}-{\frac {1}{3}}x-{\frac {1}{45}}x^{3}-{\frac {2}{945}}x^{5}-\cdots ,\qquad {\text{for }}0<|x|<\pi .\end{aligned}}}$

${\displaystyle {\begin{aligned}\cot x&=\sum _{n=0}^{\infty }{\frac {(-1)^{n}2^{2n}B_{2n}}{(2n)!}}x^{2n-1}\\[5mu]&=x^{-1}-{\frac {1}{3}}x-{\frac {1}{45}}x^{3}-{\frac {2}{945}}x^{5}-\cdots ,\qquad {\text{for }}0<|x|<\pi .\end{aligned}}}$

Parity:

The cosine and the secant are even functions; the other trigonometric functions are odd functions. That is:

${\displaystyle {\begin{aligned}\sin(-x)&=-\sin x\\\cos(-x)&=\cos x\\\tan(-x)&=-\tan x\\\cot(-x)&=-\cot x\\\csc(-x)&=-\csc x\\\sec(-x)&=\sec x.\end{aligned}}}$

${\displaystyle {\begin{aligned}\sin(-x)&=-\sin x\\\cos(-x)&=\cos x\\\tan(-x)&=-\tan x\\\cot(-x)&=-\cot x\\\csc(-x)&=-\csc x\\\sec(-x)&=\sec x.\end{aligned}}}$

Periods :

All trigonometric functions are periodic functions of period 2π. This is the smallest period, except for the tangent and the cotangent, which have π as smallest period. This means that, for every integer k, one has

${\displaystyle {\begin{aligned}\sin(x+2k\pi )&=\sin x\\\cos(x+2k\pi )&=\cos x\\\tan(x+k\pi )&=\tan x\\\cot(x+k\pi )&=\cot x\\\csc(x+2k\pi )&=\csc x\\\sec(x+2k\pi )&=\sec x.\end{aligned}}}$

${\displaystyle {\begin{aligned}\sin(x+2k\pi )&=\sin x\\\cos(x+2k\pi )&=\cos x\\\tan(x+k\pi )&=\tan x\\\cot(x+k\pi )&=\cot x\\\csc(x+2k\pi )&=\csc x\\\sec(x+2k\pi )&=\sec x.\end{aligned}}}$