Question

Sin 2theta+cos theta+tan theta sin theta+cos3 theta =sec theta

Answer 1

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By proving LHS=RHS

LHS

=> cos³∅ + cos∅ + sin²∅ + tan∅sin∅

=> cos⁴∅ + sin²∅+$\frac{sin(theeta)}{cos(theeta)}$×sin∅

=> cos⁴∅+sin²∅+sin²∅×$\frac{1}{cos(theeta)}$

=> cos⁴∅+sin⁴∅×$\frac{1}{cos(theeta)}$

=> (cos²∅+sin²∅)²×$\frac{1}{cos(theeta)}$

=> $\frac{1}{cos(theeta)}$ = sec∅

Hence, sec∅=RHS

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Answer 2

sin²θ . cosθ + tanθ . sinθ + cos³θ = secθ

LHS = sin²θ . cosθ + tanθ . sinθ + cos³θ

= (sin²θ . cosθ + cos³θ) + tanθ . sinθ

= cosθ (sin²θ + cos²θ) + sinθ . sinθ

cosθ

= cosθ + sin²θ

cosθ

= cos²θ + sin²θ

cosθ

{ ∵ sin²θ + cos²θ = 1 }

= 1

cosθ

= secθ [RHS]

LHS = RHS Hence Proved.