Question

Evaluate sin(50+theta)-cos(40-theta)÷sin40.Cosec40+tan1.Tan40.Tan50.Tan89÷4 (cos2 29+cos2 61)

Answer 1

Here your answer goes

Step :- 1

Given ,

Sin 50° can be written as 90° - 45°

⇒ Sin (90 - 40 + theta ) - Cos ( 40 - theta )

Step :- 2

⇒ sin[90-(40-theta)]- cos(40-theta)

⇒ sin[90-(40-theta)]- cos(40-theta)

⇒

∴  sin ( 90 - theta ) = Cos theta

⇒ 0

Therefore ,  Sin ( 50 + theta ) - Cos( 40 - theta ) = 0

↓↓↓

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Answer 2

Answer:

1/4

Step-by-step explanation:

sin(50+a)-cos(40-a)/sin40*cosec40+tan1*tan40*tan50*tan89/4(cos^2 29+cos^2 61)

sina=cos(90-a)

sina=1/coseca

tana=cot(90-a)

tana=1/cota

sin^2a+cos^2a=1

sin(50+a)-sin(90-(40-a)/1+tan1*cot(90-89)*tan40*cot(90-50)/4(cos^2 29+sin^2 (90-61)

sin(50+a)-sin(50+a)/1+tan1*cot1*tan40*cot40/4(cos^2 29+sin^2 29)

0/1+1*1/4(1)

0+1/4

=1/4 plzz mark it as brainliest