question 32 pls help me.....
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(tanθ+2)(2tanθ+1)=5tanθ+2sec²θ
LHS=(tanθ+2)(2tanθ+1)
tanθ(2tanθ(tanθ+2)(2tanθ+1)=5tanθ+2sec²θ+1)+2(2tanθ+1)
2tan²θ+tanθ+4tanθ+2
2tan²θ+5tanθ+2
2(sec²θ-1)+5tanθ+2
2sec²θ-2+5tanθ+2
5tanθ+2sec²θ-2+2
5tanθ+2sec²θ=RHS
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