Question

plz give me solution step by step​

Answer 1

$\huge{\mathfrak{Question:-}}$

$\large{\sf{\frac{tan\;A+tan\;B}{cot\;A+cot\;B}=tan\;A.\;tan\;B}}$

$\huge{\mathfrak{Solution:-}}$

$\large{\sf{\frac{tan\;A+tan\;B}{cot\;A+cot\;B}=tan\;A.\;tan\;B}}$

$\large{\sf{\frac{tan\;A+tan\;B}{cot\;A+cot\;B}}}$

$\large{\sf{\implies\frac{tan\;A+tan\;B}{\frac{1}{tan\;A}+\frac{1}{tan\;B}}}\;\;\;\; \Big(\because cot\;x = \frac{1}{tan\;x}\Big)}$

$\large{\sf{\implies \frac{\tan\;A+tan\;B}{\frac{tan\;A+tan\;B}{tan\;A.\;tan\;B}}}}$

$\large{\sf{\implies tan\;A.\;tan\;B}}$

$\large{\bf{Hence\;Proved}}$

$\bf{\underline{Some\;Trigonometric\;Identities:-}}$

$\sf{1).\; sin(90^{\circ}-A)=cosA}$

$\sf{2).\; tan(90^{\circ}-A)=cotA}$

$\sf{3).\; sec(90^{\circ}-A)=cosecA}$

$\sf{4).\; cos^{2}A+sin^{2}A=1}$

$\sf{5).\; 1+tan^{2}A=sec^{2}A}$

$\sf{6).\; cot^{2}A+1=cosec^{2}A}$