Verify the Identity:
(Tanx + 1)/(secx + cscx) = sinx
tan+1 = sin(sec+csc)
= sin/cos + 1
answer
To verify the given identity, we need to manipulate the left-hand side (LHS) of the equation until it matches the right-hand side (RHS) of the equation.
Let's start by simplifying the LHS of the equation.
1. Firstly, let's convert tanx, secx, and cscx into their respective trigonometric functions.
Recall the following trigonometric identities:
- tanx = sinx/cosx
- secx = 1/cosx
- cscx = 1/sinx
By substituting these values, we get:
LHS = [(sinx/cosx) + 1] / [(1/cosx) + (1/sinx)]
2. Next, let's simplify the expression further.
To combine the fractions in the denominator, we need a common denominator, which is sinx * cosx.
By multiplying the terms in the denominator by cosx/sinx and 1/cosx respectively, we can obtain the common denominator.
LHS = [(sinx + cosx) / cosx] / [(cosx * sinx / sinx * cosx) + (sinx * cosx / sinx * cosx)]
= [(sinx + cosx) / cosx] / [(sinx * cosx + sinx * cosx) / (sinx * cosx)]
= (sinx + cosx) / cosx / (2sinx*cosx) / (sinx*cosx) [Combining terms and simplifying]
= [(sinx + cosx) / cosx] * [(sinx*cosx) / (2sinx*cosx)] [Dividing by a fraction is equivalent to multiplying by its reciprocal]
= (sinx + cosx) * (sinx*cosx) / [cosx * 2sinx*cosx] [Multiplying fractions]
= (sinx + cosx) * sinx * cosx / (cosx * 2sinx * cosx)
= (sinx + cosx) * sinx / (2sinx)
= (sinx + cosx) / 2
So, LHS = (sinx + cosx) / 2
Finally, let's compare the LHS and RHS of the equation:
LHS = (sinx + cosx) / 2
RHS = sinx
After simplification, we can see that the LHS is equal to the RHS. Hence, the identity is verified.
To summarize:
(Tanx + 1)/(secx + cscx) = sinx
(sinx + cosx) / 2 = sinx