sin t 3 Solution25 sect tant2 1cost sintcost2 1sintcost2 1si

Solution

25)

(sect +tant)²

=((1_/cost) +(sint_/cost))²

=((1+sint)_/cost)²

=(1+sint)²_/cos²t

=(1+sint)²_/(1-sin²t)...........................................................since sin²t+cos²t =1 is an identity

=(1+sint)²_{/((1-sint)(1+sint))}

=(1+sint)_/(1-sint)

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30)

a³-b³=(a-b)(a²+ab+b²)

[cos³x-sin³x]_{/(cosx -sinx)}

=[(cosx-sinx)(cos²x+cosxsinx+sin²x)]_{/(cosx -sinx)}

=(cos²x+cosxsinx+sin²x)

=((cos²x+sin²x)+sinxcosx)

=1+ sinxcosx

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36)

(cscx_/(1+cscx))-(cscx_/(1-cscx))

=cscx[(1_/(1+cscx))-(1_/(1-cscx))]

=cscx[((1-cscx)-(1+cscx))_{/(1+cscx)(1-cscx)}]

=cscx[(-2cscx)_/(1-csc²x)]

=(2csc²x)_{/(csc²x -1)}

=(2csc²x)_{/(csc²x(1 -sin²x))}

=2_{/(1 -sin²x)}

=2_/(cos²x)

=2sec²x

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41)

(1-tan²)²

=(1-(sec²-1))²

=(2-sec²)²

=2²-2*2sec²+(sec²)²

=4-4sec²+sec⁴

=(4(1-sec²))+sec⁴

=(4(-tan²))+sec⁴

=-4tan²+sec⁴

=sec⁴-4tan²