I sin-49o-cos-41o 2- cos-41o-sin-49o 2 ii cos-48o-sin-42o iii cot-40o-tan-50o-1-2 cos-35o-sin-55o iv sin-27o-cos-63o 2- cos-63o-sin-27o 2 v tan-35o-tan-55o-cot-78o-tan-12o-1 vi sec-70o-cosec-20o-sin-59o-cos-31o vii cosec 31o-sec 59o viii sin-72o-cos-18osin-72o-cos-18o ix sin 35o-sin-55o-cos-35o-cos-55o x tan 48o-tan-23o-tan-42o-tan-67o xi sec-50o-sin-40o-cos-40o-cosec-50o

I) sin(90° α) = cosα and, cos(90° α) = sinα ∴(sin49°/cos41° )²+(cos41°/sin49° )² =(sin(90° 41)/cos41° )²+(cos(90° 49°)/sin49° )² =1+1 =2 ii) cos 48 sin 42=c ...

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Byju's Answer

Standard X

Mathematics

Complementary Trigonometric Ratios

i sin 49o/cos...

Question

(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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Solution

i) sin(90°-α) = cosα
and, cos(90°-α) = sinα

∴( $\frac{s i n 49 °}{c o s 41 °}$ )²+( $\frac{c o s 41 °}{s i n 49 °}$ )²
=( $\frac{s i n (90 ° - 41)}{c o s 41 °}$ )²+( $\frac{c o s (90 ° - 49 °)}{s i n 49 °}$ )²
=1+1
=2

ii) cos 48 - sin 42=cos 48-cos(90-42)

=cos 48-cos 48
=0

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

tan50 = tan(90-40) =cot40
Sin55=sin(90-35)=cos 35

= $\frac{c o t 40 °}{c o t 40 °}$ - $\frac{1}{2}$ $\frac{c o s 35 °}{c o s 35 °}$

=1- $\frac{1}{2}$
(1)

=1- $\frac{1}{2}$
= $\frac{1}{2}$

= $\frac{1}{2}$

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

=( $\frac{s i n (90 - 63) °}{c o s 63 °}$ )²+( $\frac{c o s 63 °}{s i n (90 - 63) °}$ )²
= ( $\frac{c o s 63 °}{c o s 63 °}$ )²+( $\frac{c o s 63 °}{c o s 63 °}$ )²
=1²+1²
=2

V)

$\frac{t a n 35 °}{c o t 55 °}$ + $\frac{c o t 78 °}{t a n 12 °}$ - 1

= $\frac{t a n (90 - 55) °}{c o t 55 °}$ + $\frac{c o t (90 - 12) °}{t a n 12 °}$ - 1

= $\frac{c o t 55 °}{c o t 55 °}$ + $\frac{t a n 12 °}{t a n 12 °}$ - 1

=1+1 - 1

=1

Vi)
$\frac{s e c 70 °}{c o s e c 20 °}$ + $\frac{s i n 59 °}{c o s 31 °}$
$\frac{s e c (90 - 70) °}{c o s e c 20 °}$ + $\frac{s i n (90 - 59) °}{c o s 31 °}$
$\frac{c o s e c 20 °}{c o s e c 20 °}$ + $\frac{c o s 31 °}{c o s 31 °}$
=1+1
=2

(vii)
cosec31°−sec59°
Cosec (90-59)° - sec59°
=sec 59° - sec59°
=0
(viii)

(sin 72°+cos 18°)(sin 72°−cos 18°)
=(sin(90-18)°+cos18°)(sin(90-18)°-cos18°)
=(cos18° + cos 18°) (cos18° - cos 18°)
=0
(ix)

sin 35° sin55° - cos35° cos55°

The value of sin 35° sin55° - cos35° cos55° has to be determined. Here it is not necessary to know the values of the sine or cosine of 35 or 55. Instead use the rule: sin asin b - cos acos b = -cos(a + b)

sin 35° sin55° - cos35° cos55°

=> -cos(35 + 55)°

=> -cos 90°

=> 0

The value of sin sin 35° sin55° - cos35° cos55°= 0

(x)

tan 48° tan 23° tan 42° tan 67°
= tan48°tan23°tan42°tan67°

= tan(90–42)°tan(90–67)°tan42°tan67°

=Cot42°Cot67°tan42°tan67°

= $\frac{1}{t a n 42 °}$ tan42° $\frac{1}{t a n 67 °}$ tan67°

=1

(Xi)
sec50°sin40°+cos40°cosec50°

Sec(90-40)°sin40° + cos(90-50)°cosec 50°
=cosec40°sin40° + sin30°*cosec50°
=1+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°

Without using trigonometric tables, prove that:

(i) $s i n 53^{\circ} c o s 37^{\circ} + c o s 53^{\circ} s i n 37^{\circ} = 1$

(ii) $c o s 54^{\circ} c o s 36^{\circ} - s i n 54^{\circ} s i n 36^{\circ} = 0$

(iii) $s e c 70^{\circ} s i n 20^{\circ} + c o s 20^{\circ} c o s e c 70^{\circ} = 2$

(iv) $s i n 35^{\circ} s i n 55^{\circ} - c o s 35^{\circ} c o s 55^{\circ} = 0$

(v) $(s i n 72^{\circ} + c o s 18^{\circ}) (s i n 72^{\circ} - c o s 18^{\circ}) = 0$

(vi) $t a n 48^{\circ} t a n 23^{\circ} t a n 42^{\circ} t a n 67^{\circ} = 1$

Q. Prove that

(sin 35^{0} cos 55^{0} + cos 35^{0} sin 55^{0}) + \sqrt{3} (tan 10^{0} tan 30^{0} tan 80^{0}) = 2

Q. Prove that :

(i) tan 20° tan 35° tan 45° tan 55° tan 70° = 1
(ii) sin 48° sec 42° + cos 48° cosec 42° = 2
(iii)

\frac{\sin 70 °}{\cos 20 °} + \frac{cosec 20 °}{\sec 70 °} - 2 \cos 70 ° cosec 20 ° = 0

(iv)

\frac{\cos 80 °}{\sin 10 °} + \cos 59 ° cosec 31 ° = 2

Q. Prove that :

s i n 35^{\circ} s i n 55^{\circ} - c o s 35^{\circ} c o s 55^{\circ} = 0

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