Bài 3 :a) \(\left(1-\dfrac{1}{cos\alpha}\right)\left(1+\dfrac{1}{cos\alpha}\right)+tan^2\alpha=0\)\(\Leftrightarrow\left(1-\dfrac{1}{cos^2\alpha}\right)+tan^2\alpha=0\left(1\right)\)mà \(1+tan^2\alpha=\dfrac{1}{cos^2\alpha}\Rightarrow1-\dfrac{1}{cos^2\alpha}=-tan^2\alpha\)\(\left(1\right)\Leftrightarrow-tan^2\alpha+tan^2\alpha=0\left(luôn.luôn.đúng\right)\)\(\Rightarrow dpcm\)b) \(tan\alpha+\dfrac{cos\alpha}{1+sin\alpha}=\dfrac{1}{cos\alpha}\) \(\left(cos\alpha\ne0;sin\alpha\ne-1\right)\)\(\Leftrightarrow\dfrac{tan\alpha.cos\alpha\left(1+sin\alpha\right)+cos^2\alpha-\left(1+sin\alpha\right)}{cos\alpha\left(1+sin\alpha\right)}=0\)\(\Leftrightarrow sin\alpha\left(1+sin\alpha\right)+cos^2\alpha-1-sin\alpha=0\)\(\Leftrightarrow sin\alpha+sin^2\alpha+cos^2\alpha-1-sin\alpha=0\)\(\Leftrightarrow sin\alpha-sin\alpha+sin^2\alpha+cos^2\alpha-1=0\)\(\Leftrightarrow0=0\left(luôn.luôn.đúng\right)\)\(\Rightarrow dpcm\)Câu trả lời: Bài 3 :a) \(\left(1-\dfrac{1}{cos\alpha}\right)\left(1+\dfrac{1}{cos\alpha}\right)+tan^2\alpha=0\)\(\Leftrightarrow\left(1-\dfrac{1}{cos^2\alpha}\right)+tan^2\alpha=0\left(1\right)\)mà \(1+tan^2\alpha=\dfrac{1}{cos^2\alpha}\Rightarrow1-\dfrac{1}{cos^2\alpha}=-tan^2\alpha\)\(\left(1\right)\Leftrightarrow-tan^2\alpha+tan^2\alpha=0\left(luôn.luôn.đúng\right)\)\(\Rightarrow dpcm\)b) \(tan\alpha+\dfrac{cos\alpha}{1+sin\alpha}=\dfrac{1}{cos\alpha}\) \(\left(cos\alpha\ne0;sin\alpha\ne-1\right)\)\(\Leftrightarrow\dfrac{tan\alpha.cos\alpha\left(1+sin\alpha\right)+cos^2\alpha-\left(1+sin\alpha\right)}{cos\alpha\left(1+sin\alpha\right)}=0\)\(\Leftrightarrow sin\alpha\left(1+sin\alpha\right)+cos^2\alpha-1-sin\alpha=0\)\(\Leftrightarrow sin\alpha+sin^2\alpha+cos^2\alpha-1-sin\alpha=0\)\(\Leftrightarrow sin\alpha-sin\alpha+sin^2\alpha+cos^2\alpha-1=0\)\(\Leftrightarrow0=0\left(luôn.luôn.đúng\right)\)\(\Rightarrow dpcm\)