I am working on a problem regarding Girsanov theorem and Brownian motion. Let $B$ be a standard Brownian motion. For $a,b>0$ we define stopping times $$ T_{a,b}=\inf\{t\geq0:B_t<bt-a\}$$ $$ S_a=\inf\{t\geq0:B_t=-a\}$$ and a martingale $$ M_t=\exp(B_t-\frac{1}{2}t).$$ Using Girsanov theorem I want to prove the following equality: $$ E(M_t\cdot\mathbb{1}(t<T_{a,1}))=P(t<S_a).$$ I also want to show that $$E(e^{\frac{b^2}{2}T_{a,b}})=e^{ab}.$$
So far I have proven the following two things: $$E(e^{\frac{1}{2}T_{a,1}})\leq e^a$$ $$|M_{T_{a,1}\land t}-M_{T_{a,1}}|\leq M_t\cdot 1(t<T_{a,1})+e^{\frac{1}{2}T_{a,1}}\cdot 1(t<T_{a,1})$$
I am struggling to see the connection between Girsanov theorem and what I want to prove. The equality reminds me of the definition of the new measure $Q$: $Q(A)=E_P(X\cdot1_A)$, where $E(X)=1$, which is true for $X:=M_t$, but as for $A$, $T_{a,1}$ and $S_a$ are not equal. Do I somehow need to prove that $B$ is a Brownian motion with a drift under new $P$? I am not sure how Girsanov theorem proves this equality. Would this equality also help me with the second one I want to prove?