sin theta sec^7 theta+cos theta cosec^7 thetasinθsec7θ+cosθcosec7θ
=sintheta/costheta sec^6 theta+costheta/sintheta cosec^6 theta=sinθcosθsec6θ+cosθsinθcosec6θ
=tantheta (1+tan^2theta)^3 +1/tantheta (1+cot^2theta)^3=tanθ(1+tan2θ)3+1tanθ(1+cot2θ)3
=tantheta (1+tan^2theta)^3 +1/tantheta (1+1/tan^2theta)^3=tanθ(1+tan2θ)3+1tanθ(1+1tan2θ)3
=sqrt(a/b) (1+a/b)^3 +sqrt(b/a) (1+b/a)^3=√ab(1+ab)3+√ba(1+ba)3
=sqrt(a/b)xx (a+b)^3/b^3 +sqrt(b/a) xx(a+b)^3/a^3=√ab×(a+b)3b3+√ba×(a+b)3a3
=(a+b)^3(sqrta/b^(7/2) +sqrtb/a^(7/2))=(a+b)3(√ab72+√ba72)
=((a+b)^3(a^4+b^4))/((ab)^(7/2))=(a+b)3(a4+b4)(ab)72
Q-2
Let
tanA=3 and tanB=2tanA=3andtanB=2
So A>BA>B
We are to find out
sin2(A-B)sin2(A−B)
Expanding we get
sin2(A-B)=sin2Acos2B-cos2Asin2Bsin2(A−B)=sin2Acos2B−cos2Asin2B
So we are to know
sin2A,cos2B,cos2A and sin2Bsin2A,cos2B,cos2Aandsin2B
Now sin2A=(2tanA)/(1+tan^2A)sin2A=2tanA1+tan2A
=>sin2A=(2xx3)/(1+3^2)=3/5⇒sin2A=2×31+32=35
And
sin2B=(2tanB)/(1+tan^2B)sin2B=2tanB1+tan2B
=>sin2B=(2xx2)/(1+2^2)=4/5⇒sin2B=2×21+22=45
cos2A=(1-tan^2A)/(1+tan^2A)cos2A=1−tan2A1+tan2A
=>cos2A=(1-3^2)/(1+3^2)=-4/5⇒cos2A=1−321+32=−45
And
cos2B=(1-tan^2B)/(1+tan^2B)cos2B=1−tan2B1+tan2B
=>cos2B=(1-2^2)/(1+2^2)=-3/5⇒cos2B=1−221+22=−35
Now inserting the values we get
sin2(A-B)sin2(A−B)
=sin2Acos2B-cos2Asin2B=sin2Acos2B−cos2Asin2B
=3/5xx(-3/5)-(-4/5)xx4/5=35×(−35)−(−45)×45
=-9/25+16/25=7/25=−925+1625=725
