Assertion (A): (1 − cosec² θ)(1 − sec² θ) = 1
Reason (R): 1 + tan² θ = sec² θ and 1 + cot² θ = cosec² θ
A is true, R is false
A is false, R is true
Both A and R are true
Both A and R are false

Question

Assertion (A): (1 − cosec² θ)(1 − sec² θ) = 1

Reason (R): 1 + tan² θ = sec² θ and 1 + cot² θ = cosec² θ

A is true, R is false
A is false, R is true
Both A and R are true
Both A and R are false

KnowledgeBoat · Accepted Answer

Solving L.H.S, (1 − cosec2 θ)(1 − sec2 θ) = (− cot2 θ)(− tan2 θ) = (tan2 θ) = 1 L.H.S = R.H.S Therefore, assertion (A) is true. 1 + tan2 θ = sec2 θ and 1 + cot2 θ = cosec2 θ These are standard trigonometric identities derived from the unit circle and the Pythagorean theorem. Therefore, reason (R) is true. Both A and R are true. Hence, option 3 is the correct option.

Mathematics

Assertion (A): (1 − cosec² θ)(1 − sec² θ) = 1
Reason (R): 1 + tan² θ = sec² θ and 1 + cot² θ = cosec² θ
A is true, R is false
A is false, R is true
Both A and R are true
Both A and R are false

Trigonometric Identities

Answer

Related Questions

$\sqrt{\Big(\dfrac{1 − \sin A}{1 + \sin A}\Big)}$ = ?
sec A + tan A
sec A − tan A
sec A tan A
none of these

Assertion (A): $\Big(\dfrac{1 + \tan \theta}{1 + \cot \theta} \Big)^2 = \tan^2 \theta$
Reason (R): tan² θ + sec² θ = 1
A is true, R is false
A is false, R is true
Both A and R are true
Both A and R are false

Assertion (A): If sec θ + tan θ = p, then sec θ = $\dfrac{2p}{p^2 + 1}$
Reason (R): sec² θ − tan² θ = 1
A is true, R is false
A is false, R is true
Both A and R are true
Both A and R are false

Assertion (A): For an acute angle θ, if sin θ = cos θ, then 2 sin² θ + tan² θ = 1
Reason (R): For any acute angle θ, sin (90° − θ) = cos (45° + θ)
A is true, R is false
A is false, R is true
Both A and R are true
Both A and R are false

Mathematics

Assertion (A): (1 − cosec2 θ)(1 − sec2 θ) = 1Reason (R): 1 + tan2 θ = sec2 θ and 1 + cot2 θ = cosec2 θA is true, R is falseA is false, R is trueBoth A and R are trueBoth A and R are false

Trigonometric Identities

Answer

Related Questions

(1−sin⁡A1+sin⁡A)\sqrt{\Big(\dfrac{1 − \sin A}{1 + \sin A}\Big)}(1+sinA1−sinA​)​ = ?sec A + tan Asec A − tan Asec A tan Anone of these

Assertion (A): (1+tan⁡θ1+cot⁡θ)2=tan⁡2θ\Big(\dfrac{1 + \tan \theta}{1 + \cot \theta} \Big)^2 = \tan^2 \theta(1+cotθ1+tanθ​)2=tan2θReason (R): tan2 θ + sec2 θ = 1A is true, R is falseA is false, R is trueBoth A and R are trueBoth A and R are false

Assertion (A): If sec θ + tan θ = p, then sec θ = 2pp2+1\dfrac{2p}{p^2 + 1}p2+12p​Reason (R): sec2 θ − tan2 θ = 1A is true, R is falseA is false, R is trueBoth A and R are trueBoth A and R are false

Assertion (A): For an acute angle θ, if sin θ = cos θ, then 2 sin2 θ + tan2 θ = 1Reason (R): For any acute angle θ, sin (90° − θ) = cos (45° + θ)A is true, R is falseA is false, R is trueBoth A and R are trueBoth A and R are false

Assertion (A): (1 − cosec² θ)(1 − sec² θ) = 1
Reason (R): 1 + tan² θ = sec² θ and 1 + cot² θ = cosec² θ
A is true, R is false
A is false, R is true
Both A and R are true
Both A and R are false

$\sqrt{\Big(\dfrac{1 − \sin A}{1 + \sin A}\Big)}$ = ?
sec A + tan A
sec A − tan A
sec A tan A
none of these

Assertion (A): $\Big(\dfrac{1 + \tan \theta}{1 + \cot \theta} \Big)^2 = \tan^2 \theta$
Reason (R): tan² θ + sec² θ = 1
A is true, R is false
A is false, R is true
Both A and R are true
Both A and R are false

Assertion (A): If sec θ + tan θ = p, then sec θ = $\dfrac{2p}{p^2 + 1}$
Reason (R): sec² θ − tan² θ = 1
A is true, R is false
A is false, R is true
Both A and R are true
Both A and R are false

Assertion (A): For an acute angle θ, if sin θ = cos θ, then 2 sin² θ + tan² θ = 1
Reason (R): For any acute angle θ, sin (90° − θ) = cos (45° + θ)
A is true, R is false
A is false, R is true
Both A and R are true
Both A and R are false