Question #28e6e

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Answer 1 · 2017-06-16T09:15:26

$sin theta sec^7 theta+cos theta cosec^7 theta$

$=sintheta/costheta sec^6 theta+costheta/sintheta cosec^6 theta$

$=tantheta (1+tan^2theta)^3 +1/tantheta (1+cot^2theta)^3$

$=tantheta (1+tan^2theta)^3 +1/tantheta (1+1/tan^2theta)^3$

$=sqrt(a/b) (1+a/b)^3 +sqrt(b/a) (1+b/a)^3$

$=sqrt(a/b)xx (a+b)^3/b^3 +sqrt(b/a) xx(a+b)^3/a^3$

$=(a+b)^3(sqrta/b^(7/2) +sqrtb/a^(7/2))$

$=((a+b)^3(a^4+b^4))/((ab)^(7/2))$

Q-2

Let
$tanA=3 and tanB=2$

So $A>B$

We are to find out

$sin2(A-B)$

Expanding we get

$sin2(A-B)=sin2Acos2B-cos2Asin2B$

So we are to know

$sin2A,cos2B,cos2A and sin2B$

Now $sin2A=(2tanA)/(1+tan^2A)$

$=>sin2A=(2xx3)/(1+3^2)=3/5$

And

$sin2B=(2tanB)/(1+tan^2B)$

$=>sin2B=(2xx2)/(1+2^2)=4/5$

$cos2A=(1-tan^2A)/(1+tan^2A)$

$=>cos2A=(1-3^2)/(1+3^2)=-4/5$

And

$cos2B=(1-tan^2B)/(1+tan^2B)$

$=>cos2B=(1-2^2)/(1+2^2)=-3/5$

Now inserting the values we get

$sin2(A-B)$

$=sin2Acos2B-cos2Asin2B$

$=3/5xx(-3/5)-(-4/5)xx4/5$

$=-9/25+16/25=7/25$