यदि $$ {Acos^{2}θ+Bsin^{2}θ}={sin^{2}θ(sec^{2}θ+1)\over{sec^{2}θ-1}}$$ है तो cotθ=?
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Q:
यदि $$ {Acos^{2}θ+Bsin^{2}θ}={sin^{2}θ(sec^{2}θ+1)\over{sec^{2}θ-1}}$$ है तो cotθ=?
- 1$${\sqrt{B-2\over2-A}}$$false
- 2$${\sqrt{B-1\over2-A}}$$true
- 3$${\sqrt{B-1\over A-2}}$$false
- 4$${\sqrt{2-B\over 2-A}}$$false
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