Prove that-i- cos -160-2 -960- -952- -160-cosec -160-2 -960- -952- -160-tan -160- -960-2- -952-sec- -960-2- -952-cos -952- -160-cot- -960- -952-1-ii- nbsp- cosec-90 -176- -952-cot-450 -176- -952-cosec-90 -176- -952-tan-180 -176- -952-tan-180 -176- -952-sec-180 -176- -952-tan-360 -176- -952-sec- -952-2-iii- sin-180 -176- -952- -160-cos-90 -176- -952- -160-tan-270 -176- -952- -160-cot-360 -176- -952-sin-360 -176- -952- -160-cos-360 -176- -952- -160-cosec- -952- -160-sin-270 -176- -952-1-iv- 1-cot -952-sec -960-2- -952-1-cot -952-sec -960-2- -952-2cot -952-v- tan -160-90 -176- -952- -160-sec-180 -176- -952- -160-sin- -952-sin-180 -176- -952- -160-cot-360 -176- -952- -160-cosec-90 -176- -952-1

I #160;LHS #160;= #160;cos2 #960;+ #952; #160;cosec2 #960;+ #952; #160;tan #960;2+ #952;sec #160; #960;2+ #952; #160;cos #160; ...

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Standard X

Mathematics

Complementary Trigonometric Ratios

Prove that (i...

Question

Prove that
(i) $\frac{\cos (2 π + θ) cosec (2 π + θ) \tan (π / 2 + θ)}{\sec (π / 2 + θ) cosθ \cot (π + θ)} = 1$
(ii) $\frac{cosec (90 ° + θ) + \cot (450 ° + θ)}{cosec (90 ° - θ) + \tan (180 ° - θ)} + \frac{\tan (180 ° + θ) + \sec (180 ° - θ)}{\tan (360 ° + θ) - \sec (- θ)} = 2$
(iii) $\frac{\sin (180 ° + θ) \cos (90 ° + θ) \tan (270 ° - θ) \cot (360 ° - θ)}{\sin (360 ° - θ) \cos (360 ° + θ) cosec (- θ) \sin (270 ° + θ)} = 1$
(iv) $\"{\"1 + cotθ - \sec (\frac{π}{2} + θ)\"}\" \"{\"1 + cotθ + \sec (\frac{π}{2} + θ)\"}\" = 2 cotθ$
(v) $\frac{\tan (90 ° - θ) \sec (180 ° - θ) \sin (- θ)}{\sin (180 ° + θ) \cot (360 ° - θ) cosec (90 ° - θ)} = 1$

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Solution

$(i) LHS = \frac{\cos (2 π + θ) \cos e c (2 π + θ) \tan (\frac{π}{2} + θ)}{\sec (\frac{π}{2} + θ) \cos θ \cot (π + θ)} = \frac{\cos θ cosec θ [- \cot θ]}{[- \cos e c θ] \cos θ \cot θ} = \frac{- \cos θ cosec θ \cot θ}{- cosec θ c os θ \cot θ} = 1 = RHS Hence proved .$

$(ii) LHS = \frac{cosec (90 ° + θ) + \cot (450 ° + θ)}{cosec (90 ° - θ) + \tan (180 ° - θ)} + \frac{\tan (180 ° + θ) + \sec (180 ° - θ)}{\tan (360 ° + θ) - \sec (- θ)} = \frac{cosec (90 ° + θ) + \cot (450 ° + θ)}{cosec (90 ° - θ) + \tan (180 ° - θ)} + \frac{\tan (180 ° + θ) + \sec (180 ° - θ)}{\tan (360 ° + θ) - \sec (- θ)} = \frac{cosec (90 ° + θ) + \cot (90 ° \times 5 + θ)}{cosec (90 ° - θ) + \tan (90 ° \times 2 - θ)} + \frac{\tan (90 ° \times 2 + θ) + \sec (90 ° \times 2 - θ)}{\tan (90 ° \times 4 + θ) - \sec (- θ)} = \frac{\sec θ + \cot (90 ° \times 5 + θ)}{cosec (90 ° - θ) + \tan (90 ° \times 2 - θ)} + \frac{\tan (90 ° \times 2 + θ) + \sec (90 ° \times 2 - θ)}{\tan (90 ° \times 4 + θ) - \sec (- θ)} = \frac{\sec θ - \tan θ}{\sec θ - \tan θ} + \frac{\tan θ - \sec θ}{\tan θ - \sec θ} = 1 + 1 = 2 = RHS Hence proved .$
$(iii) LHS = \frac{\sin (180 ° + θ) \cos (90 ° + θ) \tan (270 ° - θ) \cot (360 ° - θ)}{\sin (360 ° - θ) \cos (360 ° + θ) cosec (- θ) \sin (270 ° + θ)} = \frac{\sin (90 \times 2 ° + θ) \cos (90 ° \times 1 + θ) \tan (90 ° \times 3 - θ) \cot (90 ° \times 4 - θ)}{\sin (90 ° \times 4 - θ) \cos (90 ° \times 4 + θ) cosec (- θ) \sin (90 ° \times 3 + θ)} = \frac{- \sin θ [- \sin θ] \cot θ [- \cot θ]}{[- \sin θ] \cos θ [- \cos e c θ] [- \cos θ]} = \frac{\sin^{2} θ \cot^{2} θ}{\sin θ \cos e c θ \cos θ \cos θ} = \frac{\sin^{2} θ \times \frac{\cos^{2} θ}{\sin^{2} θ}}{\sin θ \times \frac{1}{\sin θ} \times \cos^{2} θ} = \frac{\cos^{2} θ}{\cos^{2} θ} = 1 = RHS Hence proved .$

$(iv) LHS = \{1 + \cot θ - \sec (\frac{π}{2} + θ)\} \{1 + \cot θ + \sec (\frac{π}{2} + θ)\} = [1 + \cot θ - \{- cosec θ\}] [1 + \cot θ + \{- cosec θ\}] = [1 + \cot θ + cosec θ] [1 + \cot θ - cosec θ] = [1 + \cot θ + cosec θ] [1 + \cot θ - cosec θ] = [\{1 + \cot (θ)\} + \{cosec θ\}] [\{1 + \cot θ\} - \{cosec θ\}] = {\{1 + \cot θ\}}^{2} - {\{cosec θ\}}^{2}$

$= 1 + \cot^{2} θ + 2 \cot θ - {cosec}^{2} θ = 2 \cot θ [∵ 1 + \cot^{2} θ = {cosec}^{2} θ] = RHS Hence proved .$

$(v) LHS = \frac{\tan (90 ° - θ) \sec (180 ° - θ) \sin (- θ)}{\sin (180 ° + θ) \cot (360 ° - θ) cosec (90 ° - θ)} = \frac{\tan (90 ° \times 1 - θ) \sec (90 ° \times 2 - θ) \sin (- θ)}{\sin (90 ° \times 2 + θ) \cot (90 ° \times 4 - θ) cosec (90 ° \times 1 - θ)} = \frac{\cot θ [- \sec θ] [- \sin θ]}{[- \sin θ] [- \cot θ] \sec θ} = \frac{\cot θ \sec θ \sin θ}{\sin θ \cot θ \sec θ} = \frac{\frac{\cos θ}{\sin θ} \times \frac{1}{\cos θ} \times \sin θ}{\sin θ \times \frac{\cos θ}{\sin θ} \times \frac{1}{\cos θ}} = \frac{1}{1} = 1 = RHS Hence proved .$

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Similar questions

Q. Prove that
(i)

\frac{\cos (2 π + θ) cosec (2 π + θ) \tan (π / 2 + θ)}{\sec (π / 2 + θ) cosθ \cot (π + θ)} = 1

(ii)

\frac{cosec (90 ° + θ) + \cot (450 ° + θ)}{cosec (90 ° - θ) + \tan (180 ° - θ)} + \frac{\tan (180 ° + θ) + \sec (180 ° - θ)}{\tan (360 ° + θ) - \sec (- θ)} = 2

(iii)

\frac{\sin (180 ° + θ) \cos (90 ° + θ) \tan (270 ° - θ) \cot (360 ° - θ)}{\sin (360 ° - θ) \cos (360 ° + θ) cosec (- θ) \sin (270 ° + θ)} = 1

(iv)

\{1 + cotθ - \sec (\frac{π}{2} + θ)\} \{1 + cotθ + \sec (\frac{π}{2} + θ)\} = 2 cotθ

(v)

\frac{\tan (90 ° - θ) \sec (180 ° - θ) \sin (- θ)}{\sin (180 ° + θ) \cot (360 ° - θ) cosec (90 ° - θ)} = 1

Prove that:

$(i) \frac{c o s (2 π + θ) c o s e c (2 π + θ) t a n (π / 2 + θ)}{s e c (π / 2 + θ) c o s θ c o t (π + θ)} = 1$

$(i i) \frac{c o s e c (90^{\circ} + θ) + c o t (450^{\circ} + θ)}{c o s e c (90^{\circ} - θ) + t a n (180^{\circ} - θ)} + \frac{t a n (180^{\circ} + θ) + s e c (180^{\circ} - θ)}{t a n (360^{\circ} + θ) - s e c (- θ)} = 2$

$(i i i) \frac{s i n (180^{\circ} + θ) c o s (90^{\circ} + θ) t a n (270^{\circ} - θ) c o t (360^{\circ} - θ)}{s i n (360^{\circ} - θ) c o s (360^{\circ} + θ) c o s e c (- θ) s i n (270^{\circ} + θ)} (i v) 1 + c o t θ - s e c (\frac{π}{2} + θ)} 1 + c o t θ + s e c (\frac{π}{2} + θ)} = 2 c o t θ$

$(v) \frac{t a n (90^{\circ} - θ) s e c (180^{\circ} - θ) s i n (- θ)}{s i n (180^{\circ} + θ) c o t (360^{\circ} - θ) c o s e c (90^{\circ} - θ)} = 1$

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. Prove that

\frac{t a n θ}{s e c θ - 1} + \frac{t a n θ}{s e c θ + 1} = 2 c o s e c θ

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