Question

plz answer correctly​

Answer 1

QUESTION 1:

Evaluate: sin 25°. cos 65° + cos 25°. sin 65°

ANSWER 1:

sin 25°. cos 65° + cos 25°. sin 65° = 1

FORMULAE:

Sin (90° - x ) = cos x

cos(90° - x ) = sin x

sin² x + cos² x = 1

EXPLANATION:

sin (90° - 65°).cos 65° + cos(90° - 65°). sin 65°

(cos 65° × cos 65°) + (sin 65° × sin 65°)

cos² 65° + sin² 65° = 1

Hence sin 25°. cos 65° + cos 25°. sin 65 = 1

QUESTION 2:

Evaluate: (1 + tan theta + sec theta).(1 + cot theta - cosec theta)

ANSWER 2:

(1 + tan theta + sec theta).(1 + cot theta - cosec theta) = 2

FORMULAE:

$\tan \theta = \frac{ \sin \theta}{ \cos \theta} \\ \\ \cot \theta = \frac{ \cos \theta}{ \sin \theta} \\ \\ \cosec \theta = \frac{1}{ \sin \theta} \\ \\ \sec \theta = \frac{1}{ \cos \theta} \\ \\ { \sin}^{2} \theta + { \cos}^{2} \theta = 1 \\ \\ ( x + y)(x - y) = {x}^{2} - {y}^{2} \\ \\ {(x + y)}^{2} = {x}^{2} + {y}^{2} + 2xy$

EXPLANATION:

$( 1 + \frac{ \sin \theta}{ \cos \theta} + \frac{1}{ \cos \theta} )( 1 + \frac{ \cos \theta}{ \sin \theta} - \frac{1}{ \sin \theta}) \\ \\ Take \: L.C.M \\ \\ ( \frac{ \cos \theta + \sin \theta + 1}{ \cos \theta} )( \frac{\sin \theta + \cos \theta - 1}{ \sin \theta } ) \\ \\ Use \: ( x + y)(x - y) = {x}^{2} - {y}^{2} \\ \\ x = \sin \theta + \cos \theta \: \: and \ \: y = 1 \\ \\ \frac{ {(\sin \theta + \cos \theta) }^{2} - {1}^{2} }{ \sin \theta \cos \theta } \\ \\ \frac{ { \sin}^{2} \theta + { \cos}^{2} \theta + 2 \sin \theta \cos \theta - 1}{ \sin \theta \cos \theta} \\ \\ \frac{1 - 1 +2\sin \theta \cos \theta }{\sin \theta \cos \theta} \\ \\ \frac{2\sin \theta \cos \theta}{\sin \theta \cos \theta} = 2$

(1 + tan theta + sec theta).(1 + cot theta - cosec theta) = 2

QUESTION 3:

If tan A = cot B , prove that: A + B = 90°

GIVEN:

tan A = cot B

TO PROVE:

A + B = 90°

FORMULAE:

tan (90° - B) = cot B

EXPLANATION:

Substitute cot B = tan (90° - B)

tan A = tan (90° - B)

Cancel tan on both sides

A = 90° - B

A + B = 90°

HENCE PROVED.

Answer 2

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