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[" 3.uf "tan9^(0)=(p)/(q)," तil "(sec^(2...

[" 3.uf "tan9^(0)=(p)/(q)," तil "(sec^(2)81^(@))/(1+cot^(2)81^(@))" GI HIT "(1)/(8)1],[[" (a) "(q)/(p)," (b) "1],[" (c) "(p^(2))/(q^(2))," (d) "(q^(2))/(p^(2))]]

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If tan9^(@)=(p)/(q) , then the value of (sec^(2)81^(@))/(1+cot^(2)81^(@)) is

If tan 9^@ = a/b , then prove that (sec^2 81^@)/(1+cot^2 81^@) = (b^2)/(a^20) .

4xx81^(-1/2)xx(81^(1/2)+81^(3/2))

sec^(2)(tan^(-1)2) +"cosec"^(2)(cot^(-1)3)=

tan9^(0)-tan27^(0)-tan63^(@)+tan81^(@)

sqrt((25)/(81)-(1)/(9))=?(2)/(3)( b) (4)/(9)( c) (16)/(81) (d) (25)/(81)

[{(-1/3)^2}^(-2)]^(-1) = (a) 1/(81) (b) 1/9 (c) -1/(81) (d) -1/9

[{(-1/3)^2}^(-2)]^(-1) = (a) 1/(81) (b) 1/9 (c) -1/(81) (d) -1/9

tan ^(-1)""(1)/(p+q)+tan ^(-1)""(q)/(p^(2)+pq+1)=cot^(-1)p