Please solve q 14 ?

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Please solve q 14 ?

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Answer 1 · 2018-05-02T18:22:07

Given that ${tan}^{2} x = 1 - a^{2}$ $tan^2x=1-a^2$

$\to sec x = \sqrt{1 + {tan}^{2} x} = \sqrt{1 + 1 - a^{2}} = \sqrt{2 - a^{2}} = {(2 - a^{2})}^{\frac{1}{2}}$ $rarrsecx=sqrt(1+tan^2x)=sqrt(1+1-a^2)=sqrt(2-a^2)=(2-a^2)^(1/2)$

$L H S = sec x + {tan}^{3} x \cdot csc x$ $LHS=secx+tan^3x*cscx$

$=secx+cancel(sinx)/cosx*(sin^2x/cos^2x)*1/cancel(sinx)$

$=secx+sec^3x*sin^2x=secx(1+sec^2x*sin^2x)$

$=secx(1+sin^2x/cos^2x)=secx*[(sin^2x+cos^2x)/cos^2x]$

$=sec^3x=((2-a^2)^((1/2)))^3=(2-a^2)^(3/2)=RHS$

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Please solve q 14 ?

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