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Question Number 56643 by Sr@2004 last updated on 20/Mar/19

Commented by Sr@2004 last updated on 20/Mar/19

$p l e a s e s o l v e 6, 7, 8, 9$
Commented by malwaan last updated on 21/Mar/19

$and 10 ?$
	Answered by tanmay.chaudhury50@gmail.com last updated on 20/Mar/19

	$c o s e c α + c o t α = 2 + \sqrt{5}$ ${c o s e c}^{2} α - {c o t}^{2} α = 1$ $(c o s e c α + c o t α) (c o s e c α - c o t α) = 1$ $c o s e c α - c o t α = \frac{1}{\sqrt{5} + 2} = \frac{\sqrt{5} - 2}{(\sqrt{5} + 2) (\sqrt{5} - 2)} = \frac{\sqrt{5} - 2}{1}$ $c o s e c α + c o t α = \sqrt{5} + 2$ $c o s e c α - c o t α = \sqrt{5} - 2$ $2 c o s e c α = 2 \sqrt{5}$ $s i n α = \frac{1}{\sqrt{5}} \to {c o s}^{2} α = 1 - \frac{1}{5} = \frac{4}{5}$ $c o s α = \frac{2}{\sqrt{5}}$
	Answered by tanmay.chaudhury50@gmail.com last updated on 20/Mar/19
	$((sin^4 θ)/x)+((cos^4 θ)/y)=(1/(x+y)) let a=sin^2 θ (a^2 /x)+(((1−a)^2 )/y)=(1/(x+y)) ya^2 +x(1−2a+a^2 )=((xy)/(x+y)) a^2 (xy+y^2 )+(1−2a+a^2 )(x^2 +xy)=xy a^2 xy+a^2 y^2 +x^2 +xy−2ax^2 −2axy+a^2 x^2 +a^2 xy=xy a^2 (xy+y^2 +x^2 +xy)−2a(x^2 +xy)+x^2 =0 a^2 (x+y)^2 −2a(x+y)x+x^2 =0 {a(x+y)−x}^2 =0 a(x+y)=x a=(x/(x+y))→sin^2 θ=(x/(x+y))[a=sin^2 θ] cos^2 θ=1−sin^2 θ=1−(x/(x+y)) cos^2 θ=(y/((x+y))) now ((sin^(2m+2) θ)/x^m )+((cos^(2m+2) θ)/y^m ) =((((x/(x+y)))^(m+1) )/x^m )+((((y/(x+y)))^(m+1) )/y^m ) =(x/((x+y)^(m+1) ))+(y/((x+y)^(m+1) )) =((x+y)/((x+y)^(m+1) )) =(1/((x+y)^m )) proved$
	$\frac{{s i n}^{4} θ}{x} + \frac{{c o s}^{4} θ}{y} = \frac{1}{x + y}$ $l e t a = {s i n}^{2} θ$ $\frac{a^{2}}{x} + \frac{{(1 - a)}^{2}}{y} = \frac{1}{x + y}$ ${y a}^{2} + x (1 - 2 a + a^{2}) = \frac{x y}{x + y}$ $a^{2} (x y + y^{2}) + (1 - 2 a + a^{2}) (x^{2} + x y) = x y$ $a^{2} x y + a^{2} y^{2} + x^{2} + x y - 2 {a x}^{2} - 2 a x y + a^{2} x^{2} + a^{2} x y = x y$ $a^{2} (x y + y^{2} + x^{2} + x y) - 2 a (x^{2} + x y) + x^{2} = 0$ $a^{2} {(x + y)}^{2} - 2 a (x + y) x + x^{2} = 0$ ${a (x + y) - x}^{2} = 0$ $a (x + y) = x$ $a = \frac{x}{x + y} \to {s i n}^{2} θ = \frac{x}{x + y} [a = {s i n}^{2} θ]$ ${c o s}^{2} θ = 1 - {s i n}^{2} θ = 1 - \frac{x}{x + y}$ ${c o s}^{2} θ = \frac{y}{(x + y)}$ $n o w$ $\frac{{s i n}^{2 m + 2} θ}{x^{m}} + \frac{{c o s}^{2 m + 2} θ}{y^{m}}$ $= \frac{{(\frac{x}{x + y})}^{m + 1}}{x^{m}} + \frac{{(\frac{y}{x + y})}^{m + 1}}{y^{m}}$ $= \frac{x}{{(x + y)}^{m + 1}} + \frac{y}{{(x + y)}^{m + 1}}$ $= \frac{x + y}{{(x + y)}^{m + 1}}$ $= \frac{1}{{(x + y)}^{m}} p r o v e d$
	Answered by tanmay.chaudhury50@gmail.com last updated on 20/Mar/19

	$t a n A = n t a n B$ $s i n A = m s i n B$ $\frac{s i n A}{s i n B} = m$ $\frac{c o s e c B}{c o s e c A} = m$ $\frac{1 + {c o t}^{2} B}{1 + {c o t}^{2} A} = m^{2} [t a n B = \frac{t a n A}{n}]$ $\frac{1 + \frac{n^{2}}{{t a n}^{2} A}}{1 + \frac{1}{{t a n}^{2} A}} = m^{2}$ $\frac{{t a n}^{2} A + n^{2}}{1 + {t a n}^{2} A} = m^{2}$ ${t a n}^{2} A + n^{2} = m^{2} + m^{2} {t a n}^{2} A$ ${t a n}^{2} A (1 - m^{2}) = m^{2} - n^{2}$ ${t a n}^{2} A = \frac{m^{2} - n^{2}}{1 - m^{2}}$ $1 + {t a n}^{2} A = \frac{m^{2} - n^{2}}{1 - m^{2}} + 1$ ${s e c}^{2} A = \frac{1 - n^{2}}{1 - m^{2}}$ ${c o s}^{2} A = \frac{1 - m^{2}}{1 - n^{2}}$ ${c o s}^{2} A = \frac{m^{2} - 2}{n^{2} - 1}$
	Answered by tanmay.chaudhury50@gmail.com last updated on 20/Mar/19

	$a s i n x = b c o s x = \frac{2 c t a n x}{1 - {t a n}^{2} x} = k (a s s u m e d)$ $a = \frac{k}{s i n x} b = \frac{k}{c o s x} c = \frac{k}{\frac{2 t a n x}{1 - {t a n}^{2} x}} \to \frac{k}{t a n 2 x}$ $\frac{a}{b} = \frac{c o s x}{s i n x} \to \frac{a^{2} - b^{2}}{a^{2} + b^{2}} = \frac{{c o s}^{2} x - {s i n}^{2} x}{{c o s}^{2} x + {s i n}^{2} x} = c o s 2 x$ $n o w \frac{{(a^{2} - b^{2})}^{2}}{a^{2} + b^{2}}$ $= (a^{2} - b^{2}) \times c o s 2 x$ $= (\frac{k^{2}}{{s i n}^{2} x} - \frac{k^{2}}{{c o s}^{2} x}) \times c o s 2 x$ $= k^{2} (\frac{{c o s}^{2} x - {s i n}^{2} x}{{s i n}^{2} {x c o s}^{2} x}) c o s 2 x$ $= \frac{4 k^{2} \times c o s 2 x \times c o s 2 x}{4 {s i n}^{2} {x c o s}^{2} x}$ $= 4 k^{2} \times \frac{{c o s}^{2} 2 x}{{(2 s i n x c o s x)}^{2}}$ $= 4 k^{2} \times \frac{{c o s}^{2} 2 x}{{s i n}^{2} 2 x}$ $= \frac{4 k^{2}}{{t a n}^{2} 2 x}$ $= 4 c^{2} [\frac{k}{t a n 2 x} = c]$
	Commented by Sr@2004 last updated on 22/Mar/19

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