Question

tan 650
cot 25°
Evaluate
cotA= tan (90° - A)
cot 25º = tan (90° -25°) = tan 65°
We know :
tan 650
cot 25°
tan 65°
= 1
tan 650
-
(1)
te: We are
sin 3A = cos (90° – 3A), we can write (1) as
(90° - 3A) =
= cos (A - 269)
COS
BC
BC
90° - 3A = A -26°
A= 29°
ich gives
erween 0 and 45°
cot 85° + cos 75º = cot (90° - 5) + cos (90° -15°)
= tan 5° + sin 15°
EXERCISE 8.3
3
Evaluate:
(iv) cosec 31°-sec 59°
tan 26
cot 64°
sin 189
cos 72°
(iii) cos 48° -sin 42°
@tan 48°tan 23 tan 42° tan 67° = 1
G) cos 38° cos 52-sin 38° sin 52° 0
Iftas 2A = cot (A - 18"), where 2A is an acute angle, find the value of A.
A =cot B, prove that A + B = 90°
If 4A - cover (A-20%), where 4 A is an acute angle. find the value of A
le lo: If sin 3A = cos (A - 26°), where 3A is an acute angle, find the value
e given that sin 3A = cos (A-26°).
90° - 3A and A-26° are both acute angles, therefore,
mple 11 : Express cot 85° + cos 75° in terms of trigonometric ratios of angles
pls answer fast having doubt ​

Answer 1

Answer:

g is the right alternative given I think so

Answer 2

Answer:

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