If $$ {1+sinθ\over {1-sinθ}}={p^{2}\over{q^{2}}}$$, then sec θ is equal to:
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Q:
If $$ {1+sinθ\over {1-sinθ}}={p^{2}\over{q^{2}}}$$, then sec θ is equal to:
- 1$$ {2p^{2}{q^{2}}\over {p^{2}+{q^{2}}}}$$false
- 2$${1\over 2}\left({q\over p}+{p\over q} \right)$$true
- 3$$ {1\over {p^{2}}}+{1\over {q^{2}}}$$false
- 4$$ {p^{2}q^{2}}\over{p^{2}+q^{2}}$$false
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