Find the value of tan15°
Answers
Answered by
33
HERE IS YOUR ANSWER..☺️☺️
.tan 15° = tan (45° - 30°)
= ( tan 45° - 30°)/(1 + tan 45° * tan30°)
= (1 - 1/√3)/(1 + 1*1/√3)
= (√3 -1 )/√3 / (√3 + 1)/√3
= (√3 -1)/ (√3 +1)
= (√3 - 1)(√3 - 1)/(√3 +1)(√3 - 1)
= (3 -2√3 +1)/ (3–1) = (4–2√3)/2
= 2 - √3
OK☺️☺️
.tan 15° = tan (45° - 30°)
= ( tan 45° - 30°)/(1 + tan 45° * tan30°)
= (1 - 1/√3)/(1 + 1*1/√3)
= (√3 -1 )/√3 / (√3 + 1)/√3
= (√3 -1)/ (√3 +1)
= (√3 - 1)(√3 - 1)/(√3 +1)(√3 - 1)
= (3 -2√3 +1)/ (3–1) = (4–2√3)/2
= 2 - √3
OK☺️☺️
sidagr04p9e7su:
aryan aese nahi nikalna ha solution chahiye
ok
tan(45°-30°)
tan(30o)=tan(2×15o)tan(30o)=tan(2×15o)
⟹13√=2tan(15o)1−tan2(15o)⟹13=2tan(15o)1−tan2(15o)
⟹3–√=1−tan2(15o)2tan(15o)⟹3=1−tan2(15o)2tan(15o)
⟹tan2(15o)+23–√tan(15o)−1=0⟹tan2(15o)+23tan(15o)−1=0
⟹tan(15o)=−23√±(23√)2−4(1)(−1)√2(1)⟹tan(15o)=−23±(23)2−4(1)(−1)2(1)
=−23√±16√2=−23±162
=−3–√±2=−3±2
So, tan(15o)=2−3–√tan(15o)=2−3 or −(2+3–√)−(2+3)
As, 15o15o falls in first quadrant, ⟹tan(15o)>0⟹tan(15o)>0
So, tan(15o)=2−3–√tan(15o)=2−3 is the solution.
⟹13√=2tan(15o)1−tan2(15o)⟹13=2tan(15o)1−tan2(15o)
⟹3–√=1−tan2(15o)2tan(15o)⟹3=1−tan2(15o)2tan(15o)
⟹tan2(15o)+23–√tan(15o)−1=0⟹tan2(15o)+23tan(15o)−1=0
⟹tan(15o)=−23√±(23√)2−4(1)(−1)√2(1)⟹tan(15o)=−23±(23)2−4(1)(−1)2(1)
=−23√±16√2=−23±162
=−3–√±2=−3±2
So, tan(15o)=2−3–√tan(15o)=2−3 or −(2+3–√)−(2+3)
As, 15o15o falls in first quadrant, ⟹tan(15o)>0⟹tan(15o)>0
So, tan(15o)=2−3–√tan(15o)=2−3 is the solution.
bhai kuch samjh me nahi aya
Similar questions