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Question

(i) (sin 49ocos 41o)2+(cos 41osin 49o)2

(ii) cos 48osin 42o

(iii) cot 40otan 50o12(cos 35osin 55o)

(iv) (sin 27ocos 63o)2+(cos 63osin 27o)2

(v) tan 35otan 55o+cot 78otan 12o1

(vi) sec 70ocosec 20o+sin 59ocos 31o

(vii) cosec31osec59o

(viii) (sin 72o+cos 18o)(sin 72ocos 18o)

(ix) sin35o sin 55o cos 35o cos 55o

(x) tan 48o tan 23o tan 42o tan 67o

(xi) sec 50o sin 40o+cos 40o cosec 50o

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Solution

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

∴(sin49°cos41° )²+(cos41°sin49°
=(sin(90°41)cos41° )²+(cos(90°49°)sin49°
=1+1
=2

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

=cos 48-cos 48
=0

iii) cot40°tan50° - 12cos35°sin55°

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

=cot40°cot40° - 12cos35°cos35°

=1-12
(1)

=1-12
=12

=12


(iv) (sin27°cos63°)²+(cos63°sin27°

=(sin(9063)°cos63°)²+(cos63°sin(9063)°
= (cos63°cos63°)²+(cos63°cos63°
=1²+1²
=2

V)

tan35°cot55°+cot78°tan12°- 1

=tan(9055)°cot55°+cot(9012)°tan12°- 1

=cot55°cot55°+tan12°tan12°- 1

=1+1 - 1

=1

Vi)
sec70°cosec20°+sin59°cos31°
sec(9070)°cosec20°+sin(9059)°cos31°
cosec20°cosec20°+cos31°cos31°
=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 a*sin b - cos a*cos 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°

=1tan42° *tan42°*1tan67°*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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