Solve cos 90- 90- tan- 90- sin 90- 90- -tan 90- - -2

LHS =cos( 90 θ #160; ) #160; ( 90 θ #160; ) #160; tanθ #160; #160; ( 90 θ #160; ) #160; sin( 90 θ #160; ) #160; ( 90 θ #160; ) #160 ...

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Byju's Answer

Standard IX

Mathematics

Trigonometric Ratios of Standard Angles

Solve cos 9...

Question

Solve $\frac{cos (90 - θ) sec (90 - θ) tan θ}{csc (90 - θ) sin (90 - θ) cot (90 - θ)} + \frac{tan (90 - θ)}{cot θ} = 2$

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Solution

LHS $= \frac{cos (90 - θ) sec (90 - θ) tan θ}{csc (90 - θ) sin (90 - θ) cot (90 - θ)} + \frac{tan (90 - θ)}{cot θ}$
As $sin A csc A = 1$ & $cos A sec A = 1$
$= \frac{tan θ}{cot (90 - θ)} + \frac{tan (90 - θ)}{cot (θ)}$
$tan θ = cot (90 - θ)$
$cot θ = tan (90 - θ)$
$= \frac{tan θ}{tan θ} + \frac{cot θ}{cot θ}$
$= 1 + 1$
$= 2$
=RHS
Hence, proved

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Similar questions

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

Q. The value of

\frac{\cos (90 ° - θ) \sec (90 ° - θ) \tan θ}{cosec (90 ° - θ) \sin (90 ° - θ) \cot (90 ° - θ)} + \frac{\tan (90 ° - θ)}{\cot θ}

is

(a) 1
(b) − 1
(c) 2
(d) −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) $s i n (50^{o} + θ) - c o s (40^{o} - θ) + t a n 1^{o} t a n 10^{o} t a n 20^{o} t a n 70^{o} t a n 80^{o} t a n 89^{o} = 1$

Q. Solve:

\frac{cos (90 + θ) sec (- θ) tan (180^{\circ} - θ)}{sec (360^{\circ} - θ) sin (180^{\circ} + θ) cot (90^{\circ} - θ)} =

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Solve cos(90−θ)sec(90−θ)tanθcsc(90−θ)sin(90−θ)cot(90−θ)+tan(90−θ)cotθ=2

LHS=cos(90−θ)sec(90−θ)tanθcsc(90−θ)sin(90−θ)cot(90−θ)+tan(90−θ)cotθAs sinAcscA=1 & cosAsecA=1=tanθcot(90−θ)+tan(90−θ)cot(θ)tanθ=cot(90−θ)cotθ=tan(90−θ)=tanθtanθ+cotθcotθ=1+1=2=RHSHence, proved

Solve $\frac{cos (90 - θ) sec (90 - θ) tan θ}{csc (90 - θ) sin (90 - θ) cot (90 - θ)} + \frac{tan (90 - θ)}{cot θ} = 2$