Find x- x 90 - - - tan 90 - - sin- - cos ec90 - - - 0

Given x(90+θ)+tan(90+θ)sinθ + (90+θ)=0 x= tan(90+θ)(tan (90+θ)sinθ+(90+θ)) x=θ( cosθ+θ) x=θ(sin^2θ/cosθ)=sinθ Hence x=sinθ ...

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Standard XII

Mathematics

Parametric Differentiation

Find x: x 9...

Question

Find x: $x cot (90 + θ) + tan (90 + θ) s i n θ + cos e c (90 + θ) = 0$

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Solution

Given $x cot (90 + θ) + tan (90 + θ) sin θ + csc (90 + θ) = 0$
$x = - tan (90 + θ) (tan (90 + θ) sin θ + csc (90 + θ))$
$x = cot θ (- cos θ + sec θ)$
$x = cot θ ({sin}^{2} θ / cos θ) = sin θ$
Hence $x = sin θ$

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Similar questions

Q. Find x from the following equations:
(i) cosec (90° + θ) + x cos θ cot (90° + θ) = sin (90° + θ)
(ii) x cot (90° + θ) + tan (90° + θ) sin θ + cosec (90° + θ) = 0

Q. Evaluate :

\frac{sin θ cos (90^{\circ} - θ) cos θ}{cosec (90^{\circ} - θ)}

-

\frac{cos θ sin (90^{\circ} - θ) sin θ}{sec (90^{\circ} - θ)}

Q. Prove the following :

\frac{cos (90^{\circ} - θ) sec (90^{\circ} - θ) tan θ}{cosec (90^{\circ} - θ) sin (90^{\circ} - θ) cot (90^{\circ} - θ)} + \frac{tan (90^{\circ} - θ)}{cot θ} = 2

Prove that:

(i) $s i n θ c o s (90^{\circ} - θ) + s i n (90^{\circ} - θ) c o s θ = 1$

(ii) $\frac{s i n θ}{c o s (90^{\circ} - θ)} + \frac{c o s θ}{s i n (90^{\circ} - θ)} = 2$

(iii) $\frac{s i n θ c o s (90^{\circ} - θ) c o s θ}{s i n (90^{\circ} - θ)} + \frac{c o s θ s i n (90^{\circ} - θ) s i n θ}{c o s (90^{\circ} - θ)} = 1$

(iv) $\frac{c o s (90^{\circ} - θ) s e c (90^{\circ} - θ) t a n θ}{c o s e c (90^{\circ} - θ) s i n (90^{\circ} - θ) c o t (90^{\circ} - θ)} + \frac{t a n (90^{\circ} - θ)}{c o t θ} = 2$

(v) $\frac{c o s (90^{\circ} - θ)}{1 + s i n (90^{\circ} - θ)} + \frac{1 + s i n (90^{\circ} - θ)}{c o s (90^{\circ} - θ)} = 2 c o s e c θ$

(vi) $\frac{s e c (90^{\circ} - θ) c o s e c θ - t a n (90^{\circ} - θ) c o t θ + c o s^{2} 25^{\circ} + c o s^{2} 65^{\circ}}{3 t a n 27^{\circ} t a n 63^{\circ}} = \frac{2}{3}$

(vii) $c o t θ t a n (90^{\circ} - θ) - s e c (90^{\circ} - θ) c o s e c θ + \sqrt{3} t a n 12^{\circ} t a n 60^{\circ} t a n 78^{\circ} = 2$

Prove the following:

(i) $s i n θ s i n (90^{o} - θ) - c o s θ c o s (90^{o} - θ) = 0$
(ii) $\frac{c o s (90^{o} - θ) s e c (90^{o} - θ) t a n θ}{c o s e c (90^{o}) s i n (90^{o} - θ) c o t (90^{o} - θ)} + \frac{t a n (90^{o} - θ)}{c o t θ} = 2$
(iii) $\frac{t a n (90^{o} - A) c o t A}{c o s e c^{2} A} - c o s^{2} A = 0$
(iv) $\frac{c o s (90^{o} - A) s i n (90^{o} - A)}{t a n (90^{o} - A)} = s i n^{2} A$
(v) $s i n (50^{o} + θ) - c o s (40^{o} - θ) + t a n 1^{o} t a n 10^{o} t a n 20^{o} t a n 70^{o} t a n 80^{o} t a n 89^{o} = 1$

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Find x: xcot(90+θ)+tan(90+θ)sinθ+cosec(90+θ)=0

Given xcot(90+θ)+tan(90+θ)sinθ+csc(90+θ)=0x=−tan(90+θ)(tan(90+θ)sinθ+csc(90+θ))x=cotθ(−cosθ+secθ)x=cotθ(sin2θ/cosθ)=sinθHence x=sinθ

Find x: $x cot (90 + θ) + tan (90 + θ) s i n θ + cos e c (90 + θ) = 0$

Given $x cot (90 + θ) + tan (90 + θ) sin θ + csc (90 + θ) = 0$
$x = - tan (90 + θ) (tan (90 + θ) sin θ + csc (90 + θ))$
$x = cot θ (- cos θ + sec θ)$
$x = cot θ ({sin}^{2} θ / cos θ) = sin θ$
Hence $x = sin θ$