Evaluate lin 49-cos 422-cos 41-sin 92

The correct option is B 2 We know sin (90 A) = cos A° ∴ ((sin (90 41)°/cos 41°)^2 + (cos (90 49) ° /sin 49 °)^2 ⇒((cos 41)°/cos 41°)^2 + (sin 49) ° /sin ...

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Standard IX

Mathematics

Variation of Trigonometric Ratios from 0 to 90 Degrees

Evaluate lin ...

Question

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

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Solution

The correct option is B
2

We know $s i n (90 - A) = c o s A °$

$∴ ((\frac{s i n (90 - 41) °}{c o s 41 °})^{2} + (\frac{c o s (90 - 49) °}{s i n 49 °})^{2}$

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

$\Rightarrow (1)^{2} + (1)^{2}$

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Similar questions

Q. Evaluate the following :

(i)

{(\frac{\sin 49 °}{\cos 41 °})}^{2} + {(\frac{\cos 41 °}{\sin 49 °})}^{2}

(ii) cos 48° − sin 42°
(iii)

\frac{\cot 40 °}{\tan 50 °} - \frac{1}{2} (\frac{\cos 35 °}{\sin 55 °})

(iv)

{(\frac{\sin 27 °}{\cos 63 °})}^{2} - {(\frac{\cos 63 °}{\sin 27 °})}^{2}

(v)

\frac{\tan 35 °}{\cot 55 °} + \frac{\cot 78 °}{\tan 12 °} - 1

(vi)

\frac{\sec 70 °}{cosec 20 °} + \frac{\sin 59 °}{\cos 31 °}

(vii) cosec 31° − sec 59°
(viii) (sin 72° + cos 18°) (sin 72° − cos 18°)
(ix) sin 35° sin 55° − cos 35° cos 55°
(x) tan 48° tan 23° tan 42° tan 67°
(xi) sec50° sin 40° + cos40° cosec 50°

(\frac{sin 49^{\circ}}{cos 41^{\circ}} - \frac{cos 17^{\circ}}{sin 73^{\circ}}) =

Q. Prove that:

(i) tan5° tan25° tan30° tan65° tan85° =

\frac{1}{\sqrt{3}}

(ii) cot12° cot38° cot52° cot60° cot78° =

\frac{1}{\sqrt{3}}

(iii) cos15° cos35° cosec55° cos60° cosec75° =

\frac{1}{2}

(iv) cos1° cos2° cos3° ... cos180° = 0
(v)

{(\frac{\sin 49 °}{\cos 41 °})}^{2} + {(\frac{\cos 41 °}{\sin 49 °})}^{2} = 2

(\frac{\sin 49 °}{\cos 41 °} - \frac{\cos 17 °}{\sin 73 °}) = ?

(a) 1
(b) 0
(c) −1
(d) Cannot be calculated

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

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Evaluate (sin 49°cos 41°)2+(cos 41°sin 49°)2

The correct option is B 2 We know sin(90−A)=cosA° ∴((sin(90−41)°cos41°)2+(cos(90−49)°sin49°)2 ⇒((cos 41)°cos 41°)2+(sin 49)°sin 49°)2 ⇒(1)2+(1)2

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

The correct option is B
2

We know $s i n (90 - A) = c o s A °$

$∴ ((\frac{s i n (90 - 41) °}{c o s 41 °})^{2} + (\frac{c o s (90 - 49) °}{s i n 49 °})^{2}$

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

$\Rightarrow (1)^{2} + (1)^{2}$