To prove the identity f(θ)≡cosθ+sinθ, we start with the given expression for f(θ):
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.
Step 2: Substitute these into the expression for f(θ).
f(θ)=1−cosθsinθcosθ+1−sinθcosθsinθ
Step 3: Simplify the denominators by finding a common denominator for each term.
1−cosθsinθ=cosθcosθ−cosθsinθ=cosθcosθ−sinθ
1−sinθcosθ=sinθsinθ−sinθcosθ=sinθsinθ−cosθ
Step 4: Substitute the simplified denominators back into the expression for f(θ).
f(θ)=cosθcosθ−sinθcosθ+sinθsinθ−cosθsinθ
Step 5: Invert the denominators and multiply.
f(θ)=cosθ⋅cosθ−sinθcosθ+sinθ⋅sinθ−cosθsinθ
f(θ)=cosθ−sinθcos2θ+sinθ−cosθsin2θ
Step 6: Notice that the denominators are negatives of each other: sinθ−cosθ=−(cosθ−sinθ).
Rewrite the second term with the common denominator cosθ−sinθ.
f(θ)=cosθ−sinθcos2θ+−(cosθ−sinθ)sin2θ
f(θ)=cosθ−sinθcos2θ−cosθ−sinθsin2θ
Step 7: Combine the fractions since they now have a common denominator.
f(θ)=cosθ−sinθcos2θ−sin2θ
Step 8: Apply the difference of squares identity, a2−b2=(a−b)(a+b), where a=cosθ and b=sinθ.
cos2θ−sin2θ=(cosθ−sinθ)(cosθ+sinθ)
Step 9: Substitute this back into the expression for f(θ).
f(θ)=cosθ−sinθ(cosθ−sinθ)(cosθ+sinθ)
Step 10: Cancel out the common term (cosθ−sinθ), assuming cosθ−sinθ=0.
f(θ)=cosθ+sinθ
Thus, we have proven that f(θ)≡cosθ+sinθ.
The final answer is cosθ+sinθ.
