Sec 41 o sin 49 o -cos49 0 - cosec41 0 - 2 - 3 tan20 o - tan 60 o - tan 70 o

sec41^∘sin49^∘+cos49^∘ cos41^∘ 2/√(3)(tan20^∘tan60^∘tan70^∘) =sec(90^∘ 49^∘)sin49^∘+cos49^∘cose(90^∘ 49^∘) 2/√(3)(tan(90^∘ 70^∘))tan60^∘tan70^∘ = ...

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Byju's Answer

Standard XII

Mathematics

Properties Derived from Trigonometric Identities

sec 41 o sin ...

Question

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

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Solution

$s e c 41^{\circ} s i n 49^{\circ} + c o s 49^{\circ} c o s 41^{\circ} - \frac{2}{\sqrt{3}} (t a n 20^{\circ} t a n 60^{\circ} t a n 70^{\circ})$
$= s e c (90^{\circ} - 49^{\circ}) s i n 49^{\circ} + c o s 49^{\circ} c o s e (90^{\circ} - 49^{\circ}) - \frac{2}{\sqrt{3}} (t a n (90^{\circ} - 70^{\circ})) t a n 60^{\circ} t a n 70^{\circ}$
$= c o s e c 49^{\circ} s i n 49^{\circ} + c o s 49^{\circ} s e c 49^{\circ} - \frac{2}{\sqrt{3}} (c o t 70^{\circ}) t a n 60^{\circ} t a n 70^{\circ}$
$= \frac{1}{s i n 49^{\circ}} s i n 49^{\circ} + c o s 49^{\circ} \frac{1}{c o s 49^{\circ}} \frac{- 2}{\sqrt{3}} (\frac{1}{t a n 70^{\circ}} . \sqrt{3} t a n 70^{\circ})$ $[∵ t a n 60^{\circ} = \sqrt{3}]$
=1+1-2
=0

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Similar questions

Q. Value of

{[\frac{sin 49^{o}}{cos 41^{o}} + \frac{cos 41^{o}}{sin 49^{o}}]}^{2}

Q. Without using trigonometric tables, evaluate the following :

sec 41^{o} . sin 49^{o} + cos 49^{o} csc 41^{o} - \frac{2}{\sqrt{3}} (tan 20^{o} . tan 60^{o} . tan 70^{o})

Using the given pattern, find the missing numbers.
$1^{2} + 2^{2} + 2^{2} = 3^{2}$
$2^{2} + 3^{2} + 6^{2} = 7^{2}$
$3^{2} + 4^{2} + 12^{2} = 13^{2}$
$4^{2} + 5^{2} + . . .^{2} = 21^{2}$
$5^{2} + . . .^{2} + 30^{2} = 31^{2}$
$6^{2} + 7^{2} + . . . .^{2} = . . . .^{2}$ .

Q. What is the stability order of the following species?

O_{2}^{2 +}

O_{2}^{+}

O_{2}

O_{2}^{-}

O_{2}^{2 -}

Using the given pattern, find the missing numbers.

1² + 2² + 2² = 3²

2² + 3² + 6² = 7²

3² + 4² + 12² = 13²

4² + 5² + _ ² = 21²

5² + _ ² + 30² = 31²

6² + 7² + _ ² = __²

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sec41osin49o+cos490.cosec410−2√3(tan20o.tan60o.tan70o)

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