Đáp án:
a) Biến đổi VT=(1+tan$\alpha$)(1+cot$\alpha$).sin$\alpha$.cos$\alpha$
= (1+tan$\alpha$ +cot$\alpha$ +tan$\alpha$.cot $\alpha$).sin$\alpha$.cos$\alpha$
= (tan$\alpha$ +cot$\alpha$ +2).sin$\alpha$.cos$\alpha$
= tan$\alpha$.sin$\alpha$.cos $\alpha$ + cot$\alpha$.sin $\alpha$.cos $\alpha$ +2.sin$\alpha$.cos $\alpha$
= $sin^{2}$$\alpha$ + $cos^{2}$$\alpha$ + 2.sin$\alpha$.cos $\alpha$
= 1 + 2.sin$\alpha$.cos $\alpha$ (1)
Biến đổi VP = (1+tan$\alpha$). $cos^{2}$$\alpha$ + (1+cot$\alpha$).$sin^{2}$$\alpha$
= $cos^{2}$$\alpha$ + tan$\alpha$.$cos^{2}$$\alpha$ + $sin^{2}$$\alpha$ + cot$\alpha$.$sin^{2}$$\alpha$
= $cos^{2}$$\alpha$ + sin$\alpha$.cos$\alpha$ + $sin^{2}$ $\alpha$ + sin$\alpha$.cos$\alpha$
= $sin^{2}$$\alpha$ + 2.sin$\alpha$.cos $\alpha$+ $cos^{2}$$\alpha$
= 1 + 2.sin$\alpha$.cos $\alpha$ (2)
Từ (1) và (2) ⇒ VT=VP (đpcm)
b) $\frac{sin\alpha}{1+cos\alpha}$ =$\frac{1-cos\alpha}{sin\alpha}$
⇔ $sin^{2}$$\alpha$ = ( 1- cos$\alpha$)(1+cos$\alpha$)
⇔ $sin^{2}$$\alpha$ = $cos^{2}$$\alpha$ - 1
⇔ $sin^{2}$$\alpha$ = $cos^{2}$$\alpha$ - ($cos^{2}$$\alpha$ + $sin^{2}$$\alpha$)
⇔ $sin^{2}$$\alpha$ = $sin^{2}$$\alpha$ (đpcm)
c) $\frac{1-cos\alpha}{sin\alpha}$.[ $\frac{(1+cos\alpha)^{2}}{sin^{2}\alpha}$-1] =2.cot$\alpha$
Ta có: $\frac{1-cos\alpha}{sin\alpha}$ = $\frac{sin\alpha}{1+cos\alpha}$ ( câu b)
Biến đổi VT = $\frac{1-cos\alpha}{sin\alpha}$.[ $\frac{(1+cos\alpha)^{2}}{sin^{2}\alpha}$-1]
= $\frac{sin\alpha}{1+cos\alpha}$.[ $\frac{(1+cos\alpha)^{2}}{sin^{2}\alpha}$-1]
= $\frac{1+cos\alpha}{sin\alpha}$ - $\frac{sin\alpha}{1+cos\alpha}$
= $\frac{1+cos\alpha}{sin\alpha}$ - $\frac{1-cos\alpha}{sin\alpha}$
= $\frac{2.cos\alpha}{sin\alpha}$ = 2.cot$\alpha$ (đpcm)
