Advertisements
Advertisements
Question
θ = 30° घेऊन खालील त्रिकोणमितीय नित्यसमानतेचा पडताळा घ्या:
1 + cot2 θ = cosec2 θ
Theorem
Advertisements
Solution
पक्ष: θ = 30°
साध्य: 1 + cot2 θ = cosec2 θ
सिद्धता:
डावी बाजू = 1 + cot2 θ
= 1 + cot2 30°
= `1 + (sqrt3)^2`
= 1 + 3
= 4 ...(i)
आता, उजवी बाजू = cosec2 θ
= cosec2 30°
= `(2/1)^2`
= 4 ...(ii)
∴ डावी बाजू = उजवी बाजू ...[समीकरण (i) आणि (ii) वरून]
∴ 1 + cot2 θ = cosec2 θ
shaalaa.com
Is there an error in this question or solution?
