If tan 40° = λ then \(\frac{tan 140° – tan 130°}{1+tan 140° tan 130°}\) = ………….
A. \(\frac{1-λ²}{λ}\)
B. \(\frac{1+λ²}{λ}\)
C. \(\frac{1+λ²}{2λ}\)
D. \(\frac{1-λ²}{2λ}\)
Answer:D
If tan 40° = λ then \(\frac{tan 140° – tan 130°}{1+tan 140° tan 130°}\) = ………….
A. \(\frac{1-λ²}{λ}\) B. \(\frac{1+λ²}{λ}\) C. \(\frac{1+λ²}{2λ}\) D. \(\frac{1-λ²}{2λ}\)