Question

Cho tam giác ABC vuông tại A < góc B=a chứng minh: a) 1+le="tan^{2}" alt="tan^{2}" align="absmiddle" class="latex-formula">a=$\frac{1}{sin^{2}a }$
làm giúp mình với

Answer 1

LHS:-

$\\ \sf\longmapsto 1+tan^2A$

$\boxed{\sf tanA=\dfrac{sinA}{cosA}}$

$\\ \sf\longmapsto 1+\dfrac{sin^2A}{cos^2A}$

$\\ \sf\longmapsto \dfrac{cos^2A+sin^2A}{cos^2A}$

$\boxed{\sf cos^2A+sin^2A=1}$

$\\ \sf\longmapsto \dfrac{1}{cos^2A}$

$\\ \sf\longmapsto \dfrac{1}{1-sin^2A}$

$\\ \sf\longmapsto \dfrac{1}{1}-\dfrac{1}{sin^2A}$

$\\ \sf\longmapsto \dfrac{1}{sin^2A}$

Hence verified