(1+cotA-cosecA) *(1+tanA+SecA)=2.Prove That R. h. s is equal to l. h. s.
Answers
Answer:
2
Step-by-step explanation:
To prove---->
-------------
(1+cotA-cosecA) (1+tanA+secA) = 2
proof ---->
---------
LHS=(1+cotA-cosecA) (1+tanA+secA)
we know that
cosA sinA 1
cotA=--------- , tanA=---------- , secA=--------
sinA cosA cosA
1
and cosecA=---------
sinA
putting these values
cosA 1 sinA 1
=(1+---------- - ------------)(1+-------- +----------)
sinA sinA cosA secA
sinA+cosA-1 cosA+sinA+1
=(----------------------------)(--------------------------)
sinA cosA
(sinA+cosA)-1 (sinA+cosA)+1
={-------------------------} {--------------------------}
sinA cosA
we have an identity
a²-b²=(a+b)(a-b) applying it here
(sinA + cosA)² - (1)²
=-----------------------------------------
sinA cosA
we have a identity
(a+b)²=a²+b²+2ab ,applying it here
sin²A + cos²A + 2sinA cosA -1
=---------------------------------------------------
sinA cosA
1 + 2 sinA cos A -1
=--------------------------------------------
sinA cosA
-1 and +1 cancel out each other
2 sinA cosA
= ------------------------------
sinA cosA
SinA and cosA cancel out each other from numerator and denominator
= 2 = RHS
Hope it helps you
Thanks for asking question and giving me to chance to answer it
Answer:
Step-by-step explanation:
(1+cot A-cosec A).(1+tanA+secA)= 2
L.H.S.
=(1+cosA/sinA-1/sinA).(1+sinA/cosA+1/cosA)
=(sinA+cosA-1)×(cosA+sinA+1)/sinA.cosA
=[(sinA+cosA)^2-(1)^2]/sinA.cosA.
=(sin^2A+cos^2A+2.sinA.cosA-1)/sinA.cosA.
=( 1+2.sinA.cosA -1)/sinA.cosA.
= 2.sinA.cosA/sinA.cosA
= 2 , proved.