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```
if alpha and beta aare the solution of the equation: atan(theta)+bsec(theta)=c find tan(alpha+beta)

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6 years ago

```							atan@+bsec@=cbsec@=c-atan@squaring bothsidesb^2sec^2@=(c-atan@)^2b^2(1+tan^2@)=c^2+a^2tan^2@-2actan@(b^2-a^2)tan^2@+2actan@+b^2-c^2=0since alpha and beta are the roots of the eq sotan(alpha)+tan(beta)=-2ac/(b^2-a^2)tan(alpha).tan(beta)=(b^2-c^2)/(b^2-a^2)tan(alpha+beta)={tan(alpha)+tan(beta)}/{1-tan(alpha).tan(beta)}=[-2ac/(b^2-a^2)]/[1-(b^2-c^2)/(b^2-a^2)]=-2ac/(c^2-a^2)=2ac/(a^2-c^2)  Ans.Thanks & RegardsRinkoo GuptaAskIITians faculty
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6 years ago
```							atan@+bsec@=cbsec@=c-atan@squaring bothsidesb^2sec^2@=(c-atan@)^2b^2(1+tan^2@)=c^2+a^2tan^2@-2actan@(b^2-a^2)tan^2@+2actan@+b^2-c^2=0since alpha and beta are the roots of the eq sotan(alpha)+tan(beta)=-2ac/(b^2-a^2)tan(alpha).tan(beta)=(b^2-c^2)/(b^2-a^2)tan(alpha+beta)={tan(alpha)+tan(beta)}/{1-tan(alpha).tan(beta)}=[-2ac/(b^2-a^2)]/[1-(b^2-c^2)/(b^2-a^2)]=-2ac/(c^2-a^2)=2ac/(a^2-c^2) Ans.Thanks & Regards
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3 years ago
```							Sir alpha and beta are roots so how did u take tan alpha and tan beta as roots...Also  how do u know that we have to square??
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3 years ago
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```
2 years ago
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