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Trigonometry Practice Test 1

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Question : 2 of 40

Marks: +1, -0

If x + y = 90°, then what is the value of

\left(1+\frac{\tan x}{\tan y}\right)\sin^2 y?

0
$\frac{1}{2}$
1
2
None of these

Solution: 👈: Video Solution

Concept:

Use complementary-angle identity: if

x+y=90^\circ

, then

\tan x=\cot y

.

Explanation:

Given

x+y=90^\circ

, so

x=90^\circ-y

.

Therefore,

\tan x=\tan(90^\circ-y)=\cot y=\frac{\cos y}{\sin y}

.

Now,

\frac{\tan x}{\tan y}=\frac{\cot y}{\tan y}=\frac{\cos^2 y}{\sin^2 y}

.

So,

1+\frac{\tan x}{\tan y}=1+\frac{\cos^2 y}{\sin^2 y}=\frac{\sin^2 y+\cos^2 y}{\sin^2 y}=\frac{1}{\sin^2 y}

.

Multiplying by

\sin^2 y

gives

\left(1+\frac{\tan x}{\tan y}\right)\sin^2 y=1

.

Answer:

\boxed{1}

(Option C)

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