Question

value of 3+cot80° cot20°/ cot80°+cot20° is equal to​

Answer 1

Given :

• 3+cot80° cot20°/ cot80°+cot20°

To find :

• it's value

Solution :

= 3 + cot80° cot20°/cot80° + cot20°

= 3sin80° sin20° - cos80° cos20°/sin80° cos20° + cos20° sin80°

= 2 sin80° sin20° + cos80° cos20° + sin80° sin20°/sin100°

= (cos60° - cos100°) + cos60°/sin100°

= 1 - cos100°/sin100°

= tan50°

therefore ,

= ø = 5π/18

Addition information :

• tanø = sinø/cosø

• secø = 1/cosø

• cotø = 1/tanø = sinø/cosø

• 1 - tan(ø/2)/1 - tan(ø/2) = ±√1 - sinø/1 + sinø

• tan ø/2 = ±√1 - cosø/1 + cosø

• sinø = Cos(90° - ø)

• cosø= sin(90° - ø)

• tanø = cot(90° - ø)

• cotø = tan(90° - ø)

• secø = cosec(90° - ø)

Answer 2

ANSWER:-

k=cot80+cot203+cot80cot20=cot(90−10)+cot(90−70)3+cot(90−10)cot(90−70)

⇒k=tan10+tan703+tan10tan70

⇒k=sin10cos70+cos10sin703cos10cos70+sin70sin10

⇒k=sin(10+70)3(2cos(10+70)+cos(10−70))+(2cos(10−70)−cos(10+70))

⇒k=sin802cos60+cos80=sin801+cos80

⇒k=2sin40cos402cos240=cot40=tan50o

Ans: B

\\ ?}^{2ANSWER

k=cot80+cot203+cot80cot20=cot(90−10)+cot(90−70)3+cot(90−10)cot(90−70)

⇒k=tan10+tan703+tan10tan70

⇒k=sin10cos70+cos10sin703cos10cos70+sin70sin10

⇒k=sin(10+70)3(2cos(10+70)+cos(10−70))+(2cos(10−70)−cos(10+70))

⇒k=sin802cos60+cos80=sin801+cos80

⇒k=2sin40cos402cos240=cot40=tan50o

Ans: B

} | [/tex]

hope it helps