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`(tantheta+2)(2 tantheta+1) = 5 tantheta+ sec^(2)theta` |
Answer» False, LHS = `(tantheta+2)(2 tantheta+1)` `=2tan^(2)theta+4 tantheta+tantheta+2` `[therefore sec^(2)theta-tan^(2)theta=1]` `=2sec^(2)theta + 5 tantheta=RHS` |
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