If $$ {tan\ θ={a\over b}}$$ then $$ {a\sinθ+b\cosθ\over{a\sinθ-b\cosθ}}$$ is
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Q:
If $$ {tan\ θ={a\over b}}$$ then $$ {a\sinθ+b\cosθ\over{a\sinθ-b\cosθ}}$$ is
- 1$$ {a\over {a^{2}+b^{2}}}$$false
- 2$$ {b\over {a^{2}+b^{2}}}$$false
- 3$$ {a^{2}-b^{2}\over {a^{2}+b^{2}}}$$false
- 4$$ {a^{2}+b^{2}\over {a^{2}-b^{2}}}$$true
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