Find tan 41 -cos 49 -

#10; #10; use> #10;cos(90^0 θ)=sinθ #10; #10; therefore>(tan41^0/cos49^0)=(tan41^0/cos(90^0 41^0)) #10; #10; =(tan41^0/sin41^ ...

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Byju's Answer

Standard X

Mathematics

Standard Values of Trigonometric Ratios

Find tan 41...

Question

Find $\frac{tan 41^{\circ}}{cos 49^{\circ}}$

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Solution

$u s e c o s (90^{0} - θ) = s i n θ$

$t h e r e f o r e (\frac{t a n 41^{0}}{c o s 49^{0}}) = (\frac{t a n 41^{0}}{c o s (90^{0} - 41^{0})})$

$= (\frac{t a n 41^{0}}{s i n 41^{0}}) = (\frac{s i n 41^{0}}{c o s 41^{0} \times sin 41^{0}})$

$= s e c 41^{0} = c o s e c 49^{0}$

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Similar questions

Q. The value of

\frac{tan 41^{\circ}}{cos 49^{\circ}}

cosec θ

, then

θ

Q. Evaluate the following:

\frac{sin 41^{\circ}}{cos 49^{\circ}}

Q. Solve:

l o g_{10} t a n 40^{\circ} l o g_{10} t a n 41^{\circ} . . . . . . l o g_{10} t a n 50^{\circ}

Without using tables, evaluate:
(i) sin $53^{\circ} \div c o s 37^{\circ}$
(ii) cos $49^{\circ} \div s i n 41^{\circ}$
(iii) tan $66^{\circ} \div c o t 24^{\circ}$

s e c 41^{o} s i n 49^{o} + c o s 49^{0} . {c o s e c 41}^{0} - \frac{2}{\sqrt{3}} ({t a n 20}^{o} . t a n 60^{o} . t a n 70^{o})

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Find tan41∘cos49∘

usecos(900−θ)=sinθ therefore(tan410cos490)=(tan410cos(900−410)) =(tan410sin410)=(sin410cos410×sin410) =sec410=cosec490

Find $\frac{tan 41^{\circ}}{cos 49^{\circ}}$

$u s e c o s (90^{0} - θ) = s i n θ$

$t h e r e f o r e (\frac{t a n 41^{0}}{c o s 49^{0}}) = (\frac{t a n 41^{0}}{c o s (90^{0} - 41^{0})})$

$= (\frac{t a n 41^{0}}{s i n 41^{0}}) = (\frac{s i n 41^{0}}{c o s 41^{0} \times sin 41^{0}})$

$= s e c 41^{0} = c o s e c 49^{0}$