To prove the identity f(θ)≡cosθ+sinθ, we start with the given expression for f(θ) and simplify it.
Given:
f(θ)=1−tanθcosθ+1−cotθsinθ
Step 1: Express tanθ and cotθ in terms of sinθ and cosθ.
We know that tanθ=cosθsinθ and cotθ=sinθcosθ.
Substitute these into the expression for f(θ):
f(θ)=1−cosθsinθcosθ+1−sinθcosθsinθ
Step 2: Simplify the denominators of each fraction.
For the first term:
1−cosθsinθ=cosθcosθ−cosθsinθ=cosθcosθ−sinθ
For the second term:
1−sinθcosθ=sinθsinθ−sinθcosθ=sinθsinθ−cosθ
Step 3: Substitute the simplified denominators back into f(θ).
f(θ)=cosθcosθ−sinθcosθ+sinθsinθ−cosθsinθ
Step 4: Invert and multiply to simplify the complex fractions.
f(θ)=cosθ⋅cosθ−sinθcosθ+sinθ⋅sinθ−cosθsinθ
f(θ)=cosθ−sinθcos2θ+sinθ−cosθsin2θ
Step 5: Make the denominators common.
Notice that sinθ−cosθ=−(cosθ−sinθ).
Rewrite the second term:
sinθ−cosθsin2θ=−(cosθ−sinθ)sin2θ=−cosθ−sinθsin2θ
Now substitute this back into the expression for f(θ):
f(θ)=cosθ−sinθcos2θ−cosθ−sinθsin2θ
Step 6: Combine the fractions.
f(θ)=cosθ−sinθcos2θ−sin2θ
Step 7: Use the difference of squares identity, a2−b2=(a−b)(a+b).
Here, a=cosθ and b=sinθ.
So, cos2θ−sin2θ=(cosθ−sinθ)(cosθ+sinθ).
Substitute this into the numerator:
f(θ)=cosθ−sinθ(cosθ−sinθ)(cosθ+sinθ)
Step 8: Cancel out the common term (cosθ−sinθ) from the numerator and denominator (assuming cosθ−sinθ=0).
f(θ)=cosθ+sinθ
Thus, we have proven that f(θ)≡cosθ+sinθ.
The final answer is cosθ+sinθ.
