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RRB Group D 1 Dec 2025 Shift 2 Solved Paper

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Question : 43 of 100

Marks: +1, -0

Find the value of

\frac{sin80^{∘}}{cos10^{∘}}-sin57^{∘}sec33^{∘}+2

.

0
2
3
1

Solution:

Concept:

Use complementary angle identities:

\sin(90^\circ - \theta) = \cos \theta

and

\sec(90^\circ - \theta) = \csc \theta

.

Explanation:

Step 1: Rewrite

\frac{\sin 80^\circ}{\cos 10^\circ}

using

\sin 80^\circ = \sin(90^\circ - 10^\circ) = \cos 10^\circ

.

Thus

\frac{\cos 10^\circ}{\cos 10^\circ} = 1

.

Step 2: Rewrite

\sin 57^\circ \sec 33^\circ

using

\sec 33^\circ = \sec(90^\circ - 57^\circ) = \csc 57^\circ

.

So

\sin 57^\circ \times \csc 57^\circ = 1

.

Step 3: Substitute into the original expression:

1 - 1 + 2 = 2

.

Therefore the value is

2

.

Answer:

2

(Option B)

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