From the 8-18- write tan x - 90- y - y - sin 90- x - cos ec 90- y - cos 90- x -

Now, nbsp;we nbsp;have:tan nbsp;x nbsp;= nbsp;ABBC, nbsp; nbsp;cot nbsp;(90 y) nbsp;= nbsp;ADDCsec nbsp;y nbsp;= nbsp;ACDC nbsp;, ...

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Standard IX

Mathematics

Introduction to Trigonometry

From the 8.18...

Question

From the 8.18, write tan x, cot (90 $-$ y), sec y, sin (90 $-$ x), cos ec(90 $-$ y), cos(90 $-$ x).

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Solution

$Now, we have : \tan x = \frac{A B}{B C}, c o t (90 - y) = \frac{A D}{D C} \sec y = \frac{A C}{D C}, \sin (90 - x) = \frac{B C}{A C} \cos e c (90 - y) = \frac{A C}{D C}, \cos (90 - x) = \frac{A B}{A C}$

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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 from following equation
i)

cos e c (90 + θ) + x c o s θ c o t (90 + θ) = s i n (90 + θ)

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. If

\frac{\cos (x - y)}{\cos (x + y)} = \frac{m}{n}

, then write the value of tan x tan y.

Q. If tan⁻¹x + tan⁻¹ y =

\frac{π}{4}

, then write the value of x + y + xy.

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From the 8.18, write tan x, cot (90-y), sec y, sin (90-x), cos ec(90-y), cos(90-x).

Now, we have:tan x = ABBC, cot (90-y) = ADDCsec y = ACDC , sin (90-x) = BCACcosec (90-y) = ACDC, cos (90-x) = ABAC

From the 8.18, write tan x, cot (90 $-$ y), sec y, sin (90 $-$ x), cos ec(90 $-$ y), cos(90 $-$ x).

$Now, we have : \tan x = \frac{A B}{B C}, c o t (90 - y) = \frac{A D}{D C} \sec y = \frac{A C}{D C}, \sin (90 - x) = \frac{B C}{A C} \cos e c (90 - y) = \frac{A C}{D C}, \cos (90 - x) = \frac{A B}{A C}$