TanA=√2-1 then find secA.sinA+tan^2A-cosecA
If you draw the triangle, you can see that
tanA = √2-1
sinA = (√2-1)/(4-2√2)
secA = (4-2√2)
So,
secA.sinA+tan^2A-cosecA
= (4-2√2) * (√2-1)/(4-2√2) + (3-2√2) - (4-2√2)/(√2-1)
= 2-3√2
To find the value of secA.sinA + tan^2A - cosecA, we first need to express all trigonometric functions in terms of tanA.
Given that tanA = √2 - 1, we can find the values of sinA, cosA, and cosecA using trigonometric identities.
We know that:
sinA = (tanA) / (secA)
cosA = (1) / (secA)
cosecA = (1) / (sinA)
Using the given value of tanA = √2 - 1, we can substitute it into these equations to find sinA, cosA, and cosecA.
sinA = (√2 - 1) / secA
cosA = 1 / secA
cosecA = 1 / sinA
Now, let's substitute these values back into the expression secA.sinA + tan^2A - cosecA.
secA.sinA + tan^2A - cosecA
= ((1 / cosA) * (sinA)) + ((tanA)^2) - (1 / sinA)
= ((1 / (1 / secA)) * ((√2 - 1) / secA)) + ((√2 - 1)^2) - (1 / ((√2 - 1) / secA))
= (secA * (√2 - 1) / secA) + ((√2 - 1)^2) - (secA / (√2 - 1))
= (√2 - 1) + (√2 - 1)^2 - (secA / (√2 - 1))
To simplify further, let's expand (√2 - 1)^2:
= (√2 - 1) + (2 - 2√2 + 1) - (secA / (√2 - 1))
= (√2 - 1) + 3 - 2√2 - (secA / (√2 - 1))
Now, let's express secA in terms of tanA:
secA = 1 / cosA
= 1 / (1 / secA)
= sec^2A
Substituting this back into the expression:
= (√2 - 1) + 3 - 2√2 - (sec^2A / (√2 - 1))
= (√2 - 1) + 3 - 2√2 - ((√2 - 1)^2 / (√2 - 1))
Expanding (√2 - 1)^2:
= (√2 - 1) + 3 - 2√2 - (2 - 2√2 + 1) / (√2 - 1))
= (√2 - 1) + 3 - 2√2 - (2√2 - 2 + 1) / (√2 - 1))
= (√2 - 1) + 3 - 2√2 - (2√2 - 1) / (√2 - 1))
Simplifying further:
= √2 - 1 + 3 - 2√2 - 2√2 + 1 / √2 - 1
= 3 - 4√2 / √2 - 1
So, the value of secA.sinA + tan^2A - cosecA is 3 - 4√2 / √2 - 1.