Evaluate the following :
(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°
Answers
(ix)SOLUTION :
Given : sin 35°sin 55°– cos 35°cos 55°
sin 35°sin 55°– cos 35°cos 55°= sin(90°− 55°)sin 55°– cos(90°−55°)cos55°
= cos 55° × sin 55° - sin 55° cos 55°
[sin(90∘− θ)= cosθ and cos(90∘−θ)=sinθ]
= 0
sin 35°sin 55°– cos 35°cos 55° = 0
Hence, the value of sin 35°sin 55°– cos 35°cos 55° is 0
(x) SOLUTION :
Given : tan 48°tan 23°tan 42°tan 67°
tan 48°tan 23°tan 42°tan 67°= tan(90°−42°)tan(90°−67°)tan 42° tan 67°
= cot 42°cot 67°tan 42°tan 67°
[tan(90∘− θ)= cot θ ]
=(tan 67°cot 67°)(tan 42°cot 42°)
= 1 × 1 =1
[tan θ × cot θ = 1]
tan 48°tan 23°tan 42°tan 67° = 1
Hence, the value of tan 48°tan23°tan42°tan 67° is 1.
(xi) SOLUTION :
Given : sec 50°sin 40°+ cos 40°cos 50°
sec 50°sin 40°+ cos 40°cos 50° = sec(90°− 40°)sin 40° + cos(90°− 50°)cosec 50°
= cosec 40° sin 40° + sin 50° cosec 50°
[cos(90∘− θ)=sin θ,sec(90∘− θ)= cosec θ]
=1 + 1 = 2
[sin θ.cosec θ =1]
sec 50°sin 40°+ cos 40°cos 50° = 2
Hence, the value of sec 50°sin 40°+ cos 40°cos 50° is 2.
HOPE THIS ANSWER WILL HELP YOU…
(ix) sin 35° sin 55° − cos 35° cos 55°
sin 35 can be written as= sin(90-55)
=cos(55)
(as 90 is the odd multiple in table 90, it will change to cos, and it is not only for 90, it is also for 270, 450.....as they are all the odd multiples of 90...., it is also positive because (90-θ) lies in the 1st quadrant, and in the first quadrant all the trigonometric ratios are positive....)
.....sin 55 can be written as sin(90-35)=
cos(35)
now, it will be regrouped as=
cos55.cos35-cos35.cos55=0
(x)tan 48° tan 23° tan 42° tan 67°=
tan (90° - 42°) tan (90° - 67°) tan 42° tan67° =
cot 42° cot 67° tan 42° tan 67° =
(cot 42° tan 42°) (cot 67° tan 67°) =
1×1 = 1
(xi)sec50° sin 40° + cos40° cosec 50°
your 11th question is in the attachment...
hope it helped you,
pls mark me as the brainliest....if my answer has helped you...
