contestada

cho tam giác ABC vuông tại A < góc B=a chứng minh: a) 1+[tex]tan^{2}[/tex]a=[tex]\frac{1}{sin^{2}a }[/tex]
làm giúp mình với

Respuesta :

LHS:-

[tex]\\ \sf\longmapsto 1+tan^2A[/tex]

[tex]\boxed{\sf tanA=\dfrac{sinA}{cosA}}[/tex]

[tex]\\ \sf\longmapsto 1+\dfrac{sin^2A}{cos^2A}[/tex]

[tex]\\ \sf\longmapsto \dfrac{cos^2A+sin^2A}{cos^2A}[/tex]

[tex]\boxed{\sf cos^2A+sin^2A=1}[/tex]

[tex]\\ \sf\longmapsto \dfrac{1}{cos^2A}[/tex]

[tex]\\ \sf\longmapsto \dfrac{1}{1-sin^2A}[/tex]

[tex]\\ \sf\longmapsto \dfrac{1}{1}-\dfrac{1}{sin^2A}[/tex]

[tex]\\ \sf\longmapsto \dfrac{1}{sin^2A}[/tex]

Hence verified