\[1) \: \: sec30+cos^2 60+tan ^2 60 = ?\]
Explanation: \[\frac{1}{2} + \left ( \frac{1}{2} \right )^2+\sqrt{3}^2 \\ = \frac{1}{2} + \frac{1}{4} + \frac{3}{1} \\ = \frac{4+2+24}{8} = \frac{30}{8} \\ = \frac{15}{4}\]
2) \[2)\: \: sec^260 + tan^230 - sec0 = ?\]
Explanation: \[22 +\left ( \frac{1}{\sqrt{3}} \right ) ^2- 1 = 4 +\frac{ 1}{3} - 1 \\= 3 + \frac{ 1}{3} = \frac{10}{3} = 3.33\]
\[3) \: \: sin90 + cos^290 + sec0 = ?\]
Explanation: \[1+0^2 + 1 = 1+1 = 2\]
4) \[4) \: \: cosec^260 - tan^230 = ?\]
Explanation: \[\left (\frac{2}{\sqrt{3}} \right )^2 - \left ( \frac{1}{\sqrt{3}} \right )^2 \\= \frac{4}{3} - \frac{1}{3} = \frac{3}{3} = 1\]
5) \[4tan^245 + 5cot^260 - 3cosec^245\]
Explanation: \[4\times 1^2 + 5\times \left ( \frac{1}{ \sqrt{3}} \right )^2 - 3 \left ( \sqrt{2} \right )^2\\ 4 \times 1+5 \times \frac{1}{3} - 3 \times 2\\ \frac{4}{1} + \frac{5}{3} - \frac{6}{1} = \frac{12+5-18}{3} = \frac{-1}{3}\]
6) \[sin30+cosec90 - cosec^230 = ?\]
Explanation: \[\frac{1}{2} + 1 - 2^2 = \frac{1}{2} + \frac{1}{1}- \frac{4}{1} \\= \frac{1+2-8}{2} = \frac{-5}{2}\]
7) \[sin30 = \frac{5}{13} तर cosec60 = ?\]
Explanation: sin30 = कोणासमोरील बाजु⁄कर्ण =5 ⁄13
Cosec60 = कर्ण ⁄कोसबा = 13 ⁄12
( ∴ कोसबा = 5, कर्ण = 13, कोलबा = 12 पायथागोरस प्रमेय)
8) \[tan60 = \frac{8}{6} तर sec60 = ?\]
Explanation:
sec60 = कर्ण⁄कोलबा = 10 ⁄6
कोसबा = 8
कोलबा = 6
कर्ण =10
9) जर cosA = 12⁄13 तर sinA ची किंमत काढा.
Explanation: cosA = कोलबा⁄कर्ण = 12⁄13
कोलबा =12
कर्ण =13
कोसबा = 5
Sina = कोलबा⁄कर्ण = 5⁄13
10) जर cot θ = 20 ⁄21 तर cosecθ ची किंमत काढा.
Explanation: costθ = कोलबा ⁄ कोसबा /= 20 ⁄ 21
cosecθ = = कर्ण ⁄ कोसबा = 29 ⁄ 21
कोलबा = 20
कोसबा = 21
कर्ण = 29
