(1 + tan 0+ sec 0) (1 + cot 0- cosec 0) =
Answers
Answer:
Rtp: (1 + tanA + secA) (1 + cotA - cosecA) = 2
On substituting the values of tanA , secA, cotA and cosecA,
(1 + sinA/cosA + 1/cosA)(1 + cosA/sinA - 1/sinA)
=[(cosA + sinA + 1)/cosA][(sinA + cosA - 1)/sinA]
=( (cosA + sinA)² - 1²)/cosAsinA
= cos²A + sin²A + 2 - 1
= 1 + 2 - 1
= 1
I hope it helped you.....
Answer:
PLS MARK BRAINLIEST
Step-by-step explanation:
Taking the LHS we first simply multiply the two brackets and get the result as follows:
1 + tan A + sec A + cot A +1 + cosec A - cosec A - sec A - 1/sinAcosA :
then the next step is to cancel the terms with the positive and negative sign and we are left with:
2 + tan A + cot A -1/sinAcosA :
then the third step is to convert the above equation into sin-cos form as follows:
2+ sinA/cosA + cosA/sinA - 1/sinAcosA :
the fourth is step is to take the LCM of the sin -cos terms leaving 2 as it is and we get:
2 + {sin[sqaure]A + cos[sqaure]A - 1}/sinAcosA :
the last and the final step is to apply the identity sin[square]A + cos[square]A=1 and then cancelling 1 from 1 we are left with 2.
hence LHS = RHS