If $$ {sin\ {θ}}={1\over {\sqrt{2}}}$$ then (tanθ+cosθ)=?
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Q:
If $$ {sin\ {θ}}={1\over {\sqrt{2}}}$$ then (tanθ+cosθ)=?
- 1$${1\over{\sqrt{2}}}$$false
- 2$${2\over{\sqrt{2}}}$$false
- 3$${3\over{\sqrt{2}}}$$false
- 4$$ {(1+\sqrt{2})\over{\sqrt{2}}}$$true
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