In triangle , the sum of the measures of angle and angle is . If , which choice is equal to ?

Solution available with step-by-step explanation

Super-Hard SAT Math Question 176 | Free SAT Prep | Aniko

00:00

Question 176·200 Super-Hard SAT Math Questions·Geometry and Trigonometry

0 / 200

In triangle $ABC$ , the sum of the measures of angle $A$ and angle $B$ is $90^\circ$ . If $\sin(A)+\sin(B)=\frac{7}{5}$ , which choice is equal to $\tan(A)+\tan(B)$ ? A⁠ni⁠kо⁠ AІ Тut⁠o‌r

When two angles add to $90^\circ$ , rewrite trig functions of one angle in terms of the other (e.g., $\sin(B)=\cos(A)$ and $\tan(B)=\cot(A)$ ). For $\tan(A)+\tan(B)$ with complementary angles, convert it to $\tan(A)+\cot(A)=\frac{1}{\sin A\cos A}$ . If you’re given a sum like $\sin A+\cos A$ , square it and use $(\sin A+\cos A)^2=1+2\sin A\cos A$ to get the product quickly, then take the reciprocal. Frοm аnіko⁠.ai

Hints

Click me!

Convert $\sin(B)$ into a function of $A$

Since $A+B=90^\circ$ , you can rewrite $B$ as $90^\circ-A$ . What is $\sin(90^\circ-A)$ in terms of $\cos(A)$ ?

Rewrite the tangent sum

Use the fact that $\tan(90^\circ-A)=\cot(A)$ , then rewrite $\tan(A)+\cot(A)$ using $\sin(A)$ and $\cos(A)$ .

Extract the product $\sin(A)\cos(A)$ from the sum $\sin(A)+\cos(A)$

Square $\sin(A)+\cos(A)$ and use $\sin^2(A)+\cos^2(A)=1$ to isolate $\sin(A)\cos(A)$ .

Finish with a reciprocal

Once you have $\sin(A)\cos(A)$ , use the expression you found for $\tan(A)+\tan(B)$ to finish. Р⁠rοp‌е⁠rtу оf ‌Аnі‌ko.аі

Desmos Guide

Enter the given sum

In Desmos, define

S = 7/5

Compute the product $\sin(A)\cos(A)$ from the identity

Use $(\sin A+\cos A)^2=1+2\sin A\cos A$ by defining

P = (S^2 - 1)/2

Evaluate the target expression

Since $\tan(A)+\tan(B)=1/(\sin A\cos A)$ , enter

E = 1/P

Match to an answer choice

Compare the exact value of E to the answer choices to select the match. Ѕourсe: aniкο.ai

Step-by-step Explanation

Use the complementary-angle relationship

Since $A+B=90^\circ$ , angles $A$ and $B$ are complementary, so

\tan(B)=\tan(90^\circ-A)=\cot(A).

Therefore,

\tan(A)+\tan(B)=\tan(A)+\cot(A).

Rewrite the sum in terms of $\sin(A)$ and $\cos(A)$

Rewrite using sine and cosine: а⁠n‍іko.‌a‌і⁠/‌ѕa⁠t

\tan(A)+\cot(A)=\frac{\sin A}{\cos A}+\frac{\cos A}{\sin A} =\frac{\sin^2 A+\cos^2 A}{\sin A\cos A} =\frac{1}{\sin A\cos A}.

Use the given information to find $\sin(A)\cos(A)$

Because $B=90^\circ-A$ , we have $\sin(B)=\cos(A)$ . So the given condition

\sin(A)+\sin(B)=\frac{7}{5}

becomes

\sin A+\cos A=\frac{7}{5}.

Square both sides:

(\sin A+\cos A)^2=\sin^2 A+2\sin A\cos A+\cos^2 A=1+2\sin A\cos A.

\left(\frac{7}{5}\right)^2=1+2\sin A\cos A \quad\Rightarrow\quad \frac{49}{25}=1+2\sin A\cos A \quad\Rightarrow\quad \sin A\cos A=\frac{12}{25}.

Compute $\tan(A)+\tan(B)$ and match a choice

From Step 2,

\tan(A)+\tan(B)=\frac{1}{\sin A\cos A}=\frac{1}{\frac{12}{25}}=\frac{25}{12}.

So the correct choice is $\frac{25}{12}$ .

Strategy

Hints

Desmos Guide

Step-by-step Explanation

Study Hints

Strategy

Hints

Desmos Guide

Step-by-step Explanation