Find x from following equationi cos ec90 - - - xcos- cot90 - - - sin90 - -

(90+θ )+xcosθ (90+θ )=sin (90+θ ) θ +xcosθtanθ =cosθ xsinθ =cosθ θ x=θ θθ ...

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Byju's Answer

Standard XII

Mathematics

Quadratic Formula for Finding Roots

Find x from f...

Question

Find x from following equation
i) $cos e c (90 + θ) + x c o s θ c o t (90 + θ) = s i n (90 + θ)$

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Solution

$csc (90 + θ) + x cos θ cot (90 + θ) = sin (90 + θ) sec θ + x cos θ tan θ = cos θ x sin θ = cos θ - sec θ x = cot θ - csc θ sec θ$

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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

Prove the identity $[cosec (90^{\circ} - θ) - sin (90^{\circ} - θ)] [cosec θ - sin θ] [tan θ + cot θ] = 1$

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

Q. Find x:

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

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 from following equationi) cosec(90+θ)+xcosθcot(90+θ)=sin(90+θ)

csc(90+θ)+xcosθcot(90+θ)=sin(90+θ)secθ+xcosθtanθ=cosθxsinθ=cosθ−secθx=cotθ−cscθsecθ

Find x from following equation
i) $cos e c (90 + θ) + x c o s θ c o t (90 + θ) = s i n (90 + θ)$

$csc (90 + θ) + x cos θ cot (90 + θ) = sin (90 + θ) sec θ + x cos θ tan θ = cos θ x sin θ = cos θ - sec θ x = cot θ - csc θ sec θ$