Question

If sin x / a = cos x / b = tan x / c = k then

A. bc + 1/ck + ak/( 1 + bk ) = 1/k( a + 1/a )
B. a² + b² + c² = 1/b²k⁴
C. bc + 1/ck + ak/( 1 + bk ) = a( 1/k + k )
D. a² + b² + c² = 1/b²k²

Answer 1

Answer:

Answer:–

$bc + \frac{1}{ck} + \frac{ak}{ 1 + bk} = \frac{1}{k} (a + \frac{1}{a} )$

Given:–

$\frac{sinx}{a} = \frac{cosx}{b} = \frac{tanx}{c} = k$

Explanation:–

$ak = sinx \\ bk = cosx \\ ck = tanx$

Now,

$bc = \frac{sinx}{ {k}^{2} }$

⇒ bc+1/ck+ak/1+bk

⇒ sinx/k²+1/tanx+sinx/1+cosx

⇒ sinx/k²+cosx/sinx+sinx-sinx.cosx/sin²x

⇒ sinx/k²+cosx/sinx+1-cosx/sinx

⇒ sinx/k²+1/sinx

⇒ a/k+1/ak

⇒ 1/k(a+1/a)

a²=sin²x/k²

b²=cos²x/k²

c²=tan²x/k²

Now,

⇒ a²+b²+c²

⇒ sin²x/k²+cos²x/k²+tan²x/k²

⇒ 1/k²(1+tan²x)

⇒ 1/k²(sec²x)

⇒ 1/k²(1/b²k²)

⇒ 1/k⁴b²

a²+b²+c²=1/b²k⁴

Answer 2

AnswEr:-

Given :

$\sf \implies \frac{sinx}{a} = \frac{cosx}{b} = \frac{tanx}{c}= k$

According to the question :

$\begin{gathered} \sf ak = \sin(x) \\\ bk = \cos(x) \\ \\ ck = \tan(x) \\ \\ bc {k}^{2} = \sin(x)\end{gathered}$

Therefore,

$\implies \sf \frac{ \sin(x) }{ {k}^{2} } + \frac{1}{tanx} + \frac{sinx}{1 + cosx}$

$\implies \sf \: \frac{sinx}{ {k}^{2} } + \frac{cosx}{sinx} + \frac{sinx}{1 + cosx}$

$\implies \sf \: \frac{sinx}{ {k}^{2} } + \frac{1}{sinx}$

$\\ \\ \implies \sf \: \frac{ak}{ {k}^{2} } + \frac{1}{ak}$

$\\ \implies \sf \boxed{ \frac{1}{k} (a + \frac{1}{a} )}$

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We know,

$\sf ak = \sin(x) \\ \\ bk = \cos(x) \\ \\ ck = \tan(x) \\\\ bc {k}^{2} = \sin(x)$

$\implies \sf \frac{sinx}{cosx} = tanx \\ \\ \implies \sf \: \frac{ak}{bk} = ck \\ \\ \\ \implies \tt \: ck = \frac{a}{b}$

Also,

$\\\\ \bigstar \: \: \sf {sin}^{2}x + {cos}^{2}x = 1$

$\implies \sf \: {a}^{2} {k}^{2} + {b}^{2} {k}^{2} = 1 \\ \\ \\ \implies \sf {a}^{2} + {b}^{2} = \frac{1}{ {k}^{2} }$

Now,

$\implies \sf {a}^{2} + {b}^{2} + {c}^{2} = ( {a}^{2} + {b}^{2} ) + ( \frac{a}{bk} ) {}^{2}$

$\implies \sf \: \frac{1}{ {k}^{2} } + \frac{ {a}^{2} }{ {b}^{2} {k}^{2} }$

$\implies \sf \frac{ {a}^{2} + {b}^{2} }{ {b}^{2} {k}^{2} }$

Substituting the value of $\sf a^2 + b^2$ ,

$\implies \sf \: \frac{\frac{1}{ {k}^{2} } }{ {b}^{2} {k}^{2} } \\ \\ \\ \implies \sf \: \boxed{ {a}^{2} + {b}^{2} + {c}^{2} = \frac{1}{ {b}^{2} {k}^{4} } }$

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