Question

প্রমাণ কর যে, $\frac{tanA}{1-cotA}+\frac{cotA}{1-tanA}$$=secAcosecA+1$

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নবম-দশম শ্রেণির গণিত পাঠ্য বইয়ের ত্রিকোণমিতিক অনুপাত (৯.১) এর সকল সমাধান পেতে এখানে ক্লিক করুন।

Answer 1

Left Hand Side

$=\frac{tanA}{1-cotA}+\frac{cotA}{1-tanA}$

$=\left(\frac{\frac{sinA}{cosA}}{1-\frac{cosA}{sinA}}\right)+$$\left(\frac{\frac{cosA}{sinA}}{1-\frac{sinA}{cosA}}\right)$

$=\left(\frac{\frac{sinA}{cosA}}{\frac{sinA-cosA}{sinA}}\right)+$$\left(\frac{\frac{cosA}{sinA}}{\frac{cosA-sinA}{cosA}}\right)$

$=\left(\frac{sinA}{cosA} \times \frac{sinA}{sinA-cosA}\right)+$$\left(\frac{cosA}{sinA} \times \frac{cosA}{cosA-sinA}\right)$

$=\frac{sin^2A}{cosA(sinA-cosA)}+$$\frac{cos^2A}{sinA(cosA-sinA)}$

$=\frac{sin^2A}{cosA(sinA-cosA)}-$$\frac{cos^2A}{sinA(sinA-cosA)}$

$=\frac{sin^3A-cos^3A}{sinA cosA (sinA-cosA)}$

$=\frac{(sinA-cosA)(sin^2A+sinA cosA+cos^2A)}{sinA cosA (sinA-cosA)}$

$=\frac{(sin^2A+sinA cosA+cos^2A)}{sinA cosA }$

$=\frac{sin^2A+cos^2A+sinA cosA}{sinA cosA }$

$=\frac{1+sinA cosA}{sinA cosA }$

$=\frac{1}{sinA cosA }+\frac{sinA cosA}{sinA cosA }$

$=\frac{1}{sinA cosA }+1$

$=\frac{1}{sinA} \cdot \frac{1}{cosA }+1$

$=cosecA \cdot secA+1$

$=secAcosecA+1$

$=$ Right Hand Side
[ Proved ]

Comments

Rules Applied :

• $tan \theta = \frac{sin \theta}{cos \theta}$
• $cot \theta = \frac{cos \theta}{\sin \theta}$
• $a^3-b^3=(a-b)(a^2+ab+b^2)$
• $sin^2\theta+cos^2\theta=1$
• $cosec \theta = \frac{1}{sin \theta}$
• $sec \theta = \frac{1}{cos \theta}$