Sin 49-tan 11-sec 41-tan 79-

The correct option is C 1⇒ sin 49^∘. sec 41^∘. tan 11^∘. tan 79^∘ = sin 49^∘. sec (90 49^∘). tan 11^∘. tan (90 11^∘) =sin 49^∘. cosec 49^∘. tan 11^∘. cot 1 ...

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Standard X

Mathematics

Trigonometric Ratios of Complementary Angles

sin 49∘.tan 1...

Question

$s i n 49^{\circ} . t a n 11^{\circ} . s e c 41^{\circ} . t a n 79^{\circ} =$ ________.

\frac{1}{2}

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Solution

The correct option is C $1$
$\Rightarrow s i n 49^{\circ} . s e c 41^{\circ} . t a n 11^{\circ} . t a n 79^{\circ}$

$= s i n 49^{\circ} . s e c (90 - 49^{\circ}) . t a n 11^{\circ} . t a n (90 - 11^{\circ})$

$= s i n 49^{\circ} . c o s e c 49^{\circ} . t a n 11^{\circ} . c o t 11^{\circ}$

$= s i n 49^{\circ} . \frac{1}{s i n 49^{\circ}} . t a n 11^{\circ} . \frac{1}{t a n 11^{\circ}}$

$= 1$

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Similar questions

Q. Prove that

s i n 49^{\circ} . t a n 11^{\circ} . s e c 41^{\circ} . t a n 79^{\circ} = 1

[2 MARKS]

Q. Prove that

s i n 49^{\circ} . t a n 11^{\circ} . s e c 41^{\circ} . t a n 79^{\circ} = 1

[2 MARKS]

(i) ${(\frac{s i n 49^{o}}{c o s 41^{o}})}^{2} + {(\frac{c o s 41^{o}}{s i n 49^{o}})}^{2}$

(ii) $c o s 48^{o} - s i n 42^{o}$

(iii) $\frac{c o t 40^{o}}{t a n 50^{o}} - \frac{1}{2} (\frac{c o s 35^{o}}{s i n 55^{o}})$

(iv) ${(\frac{s i n 27^{o}}{c o s 63^{o}})}^{2} + {(\frac{c o s 63^{o}}{s i n 27^{o}})}^{2}$

(v) $\frac{t a n 35^{o}}{t a n 55^{o}} + \frac{c o t 78^{o}}{t a n 12^{o}} - 1$

(vi) $\frac{s e c 70^{o}}{c o s e c 20^{o}} + \frac{s i n 59^{o}}{c o s 31^{o}}$

(vii) $c o s e c 31^{o} - s e c 59^{o}$

(viii) $(s i n 72^{o} + c o s 18^{o}) (s i n 72^{o} - c o s 18^{o})$

(ix) $s i n 35^{o} s i n 55^{o} - c o s 35^{o} c o s 55^{o}$

(x) tan $48^{o} t a n 23^{o} t a n 42^{o} t a n 67^{o}$

(xi) $s e c 50^{o} s i n 40^{o} + c o s 40^{o} c o s e c 50^{o}$

Q. If

y = \tan^{- 1} (\frac{2 x}{1 - x^{2}}) + \sec^{- 1} (\frac{1 + x^{2}}{1 - x^{2}}), x > 0

, prove that

\frac{d y}{d x} = \frac{4}{1 + x^{2}} .

Q. The integral

\int \frac{{sec}^{2} x}{(sec x + tan x)^{9 / 2}} d x

equals

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sin 49∘.tan 11∘.sec 41∘.tan 79∘=________.

The correct option is C 1⇒sin 49∘.sec 41∘.tan 11∘.tan 79∘ =sin 49∘.sec (90−49∘).tan 11∘.tan (90−11∘) =sin 49∘.cosec 49∘.tan 11∘.cot 11∘ =sin 49∘.1sin 49∘.tan 11∘.1tan 11∘ =1

$s i n 49^{\circ} . t a n 11^{\circ} . s e c 41^{\circ} . t a n 79^{\circ} =$ ________.