Sin A - sin 90- A -cos A - cos 90- A - Acosec A

Given that sinA/sin(90 A) +cosA/cos(90 A) sinA/cosA+cosA/sinA (sin(90 A)=cosA on taking lcm sinA.sinA+cosA.cosA/sinA.cosA (sin ...

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Standard X

Mathematics

Standard Values of Trigonometric Ratios

Sin A / sin 9...

Question

$\frac{sin A}{sin (90 - A)} + \frac{cos A}{cos (90 - A)} = s e c A c o s e c A$

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Solution

We know that, $sin (90 - θ) = cos θ$ and $cos (90 - θ) = sin θ$
$∴ \frac{sin A}{sin (90 - A)} + \frac{cos A}{cos (90 - A)} = \frac{sin A}{cos A} + \frac{cos A}{sin A}$
$= \frac{{sin}^{2} A + {cos}^{A}}{cos A sin A}$
$= \frac{1}{cos A sin A}$ [ $∵ {cos}^{2} A + {sin}^{2} A = 1$ ]
$= sec A cosec A$
Hence,proved

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Q. Prove that:

\frac{cos A}{sin (90^{\circ} - A)} + \frac{sin A}{cos (90^{\circ} - A)} = 2

Q. Prove that

\frac{cos 9^{0} + sin 9^{0}}{cos 9^{0} - sin 9^{0}} = cot 36^{0}

Sin(45^o – A) cos(45^o - B) + cos(45^o – A) sin(45^o – B)

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

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

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sin Asin (90−A)+cos Acos (90−A)=sec Acosec A

We know that, sin(90−θ)=cosθ and cos(90−θ)=sinθ∴sin Asin (90−A)+cos Acos (90−A)=sin AcosA+cos AsinA=sin2 A+cosAcosAsinA=1cosAsinA [∵cos2A+sin2A=1]=sec A cosec AHence,proved

$\frac{sin A}{sin (90 - A)} + \frac{cos A}{cos (90 - A)} = s e c A c o s e c A$