Question

(1+tan10°)(1+tan20°)(1+tan25°)(1+tan35°)=4

Answer 1

Step-by-step explanation:

1+tan10°=sin10°+cos10°cos10°=(2–√)(sin(10°+45°)cos10°)1+tan⁡10°=sin⁡10°+cos⁡10°cos⁡10°=(2)(sin⁡(10°+45°)cos⁡10°)

where sin(10°+45°)=(sin10°/2–√)+(cos10°/2–√)sin⁡(10°+45°)=(sin⁡10°/2)+(cos⁡10°/2) from the formula for the sine of a sum. Then, continuing:

1+tan10°=(2–√)(sin(10°+45°)cos10°)=(2–√)(cos35°cos10°)1+tan⁡10°=(2)(sin⁡(10°+45°)cos⁡10°)=(2)(cos⁡35°cos⁡10°)

using sin(10°+45°)=cos(90°−10°−45°)=cos35°sin⁡(10°+45°)=cos⁡(90°−10°−45°)=cos⁡35°. Do the same with arguments of 20°,25°,35°20°,25°,35° in place of 10°10° and multiply the four resulting fractions together; all the trig functions cancel out of the product and you have just (2–√)4=4(2)4=4.

The business with A,B,CA,B,C, however, has me completely stumped. It does not enter the above equality at all!

Answer 2

Answer:: (1+ tan) (1+tan20) (1+tan 25) (1+ tan 35)=4

[(1+tan 10) (1+tn 35) (1+tan20) (I+ tan 25) = 4

(1 + tan10 + tan35 + tan tan35) (1+ tan 20+ tan 25+ tan 20⋅ tan 25)=4...

+Y= 45

tan+ tan Y/1-tanXtany=tan 45

tan X+tan Y= 1-tan. tan

tan n + tan Y + tan n. tan Y = 1

tan + tany + tan n. tany = 1

Now, Substituting value of 2 and Y in egn i in respective rea.

(1+tan10 + tan35+ tan10 tan35) (1+ tan 25+ tan 20+ tan 20. tan 25) = 4 (1+1)(1+1)