Prove the identity 1. cot²x= csc²x-1 2. sin x tan x= sec x - cos x
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The triangle ABC is shifted upwards by 3 units and towards the left by 4 units. After translation again it is rotated 90° counterclockwise. All these transformations are shown in the graph below.
<h3>What are transformation rules?</h3>The following are the basic transformation rules:
- Translation: (x, y) → (x + a, y + b) or (x, y) → (x - a, y - b)
- Reflection over x-axis: (x, y) → (x, -y)
- Reflection over y-axis: (x, y) → (-x, y)
- Rotation 90°(counterclockwise): (x, y) → (-y, x)
- Rotation 180°: (x, y) → (-x, -y)
A triangle ABC is given in the graph with vertices as
A(-1, 2), B(1, 4), and C(4, -1)
The transformation to be applied is A"B"C" = Ro 90°(T(-4,3)(ABC))
Where -
T(-4, 3) represents the translation of 3 units upwards and 4 units towards the left
R0 90° represents the rotation of 90° (counterclockwise) of the translated triangle.
Step 1: applying translation (-4, 3):
The new vertices of the triangle ABC translated by (-4, 3) units are represented by A'B'C'. They are:
A' - (-1 - 4, 2 + 3) = (-5, 5)
B' - (1 - 4, 4 + 3) = (-3, 7)
C' - (4 - 4, -1 + 3) = (0, 2)
These are shown in the graph below.
Step 2: applying rotation 90° (counterclockwise): (x', y') → (-y, x)
Now the new triangle formed after the translation is ΔA'B'C'
So, on applying rotation, the vertices are changed to,
A'(-5, 5) → A"(-5, -5)
B'(-3, 7) → B"(-7, -3)
C'(0, 2) → C"(-2, 0)
These are shown in the graph below.
Therefore, Δ ABC → Δ A'B'C' (by T(-4, 3)) → Δ A"B"C" (by Ro 90°) i.e.,
Δ A"B"C" = Ro 90°(T(-4, 3) ABC).
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Step 1. Simplify 
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Step 2. Simplify
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Step 3. Cancel
Done! :) Your answer is
Step 2. Simplify
Step 3. Cancel
Done! :) Your answer is