1. (1 – cos²0)csc²0 = 1 2. tane sino + cose = sece 3. (1– cos0)(1 + cos0)(1 + cot²0) = 1 %3D cotA + tanA = secA cscA

Q: -p² sin(w) cos?(w) – p cos² (») -p2 sin(4).

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A: We have to solve the problem

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A: Given:

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Q: The amplitude of y = } cos 3x is and the %3D period is .

A: Given: y=12cos3x To Find: Amplitude and Period Concept: if y=AcosBxthen Amplitude=Aand Period=2πB.

Q: evaluate the following giving exact values when possible: cos(sin^(-1)(y/2))

A: Simplifying the given inverse trigonometric expression to find the desired value.

Q: 31T tan 2. (")- sec(-- 4 20n cot(- + csc( 3 8

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A:

Q: 8.) k(x)=-2 tan(2(x+)) Period = Increment =

A: Given function is: kx=-2tan2x+π8 Then the period of this function is 2π2=π. The increment is π8.

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A: TO SHOW 1−cosx1+cosx=sin2x

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A: a) We know that Therefore, we have

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Q: 4л coS 씀Cos 즙Cs -sin sin 15 15 15

A: We have to find the exact value cosπ15cos4π15-sinπ15sin4π15.

Q: 7. Verify the following: (sec0 + csc®)(cos0 – sin0) = cot0 – tan0

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Q: If cos x = 4/5, csc x < 0, then sin 2x cos2x = tan 2x

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Q: ) cos(cos-¹(35)) ) sec(tan-¹(5)) csc(sin-¹(n)) ) cos ¹(cos(29°)) (4π) 5 =) sin ¹[sin ]]

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Q: w/N Find sin2x and cos2x given cot x x in QIII.

A: