Tangent, Cotangent, Secant, Cosecant in Terms of Sine and Cosine
Trending Questions
Q.
If are the smallest positive angles in ascending order of magnitude which have their sines equal to the positive quantity , then the value of is equal to
None of these
Q. The numerical value of (1+cotx−cosec x)(1+tanx+secx) is
Q.
How do you find the exact values of ?
Q. The value of sin75∘+cos75∘ is
- √3√2
- 1√2
- √3−1√2
- √3+1√2
Q. If θ∈(0, π4) and t1=(tanθ)tanθ, t2=(tanθ)cotθ, t3=(cotθ)tanθ and t4=(cotθ)cotθ then
- t1>t2>t3>t4
- t4>t3>t1>t2
- t3>t1>t2>t4
- t2>t3>t1>t2
Q. If θ∈(0, π4) and t1=(tanθ)tanθ, t2=(tanθ)cotθ, t3=(cotθ)tanθ and t4=(cotθ)cotθ then
- t1>t2>t3>t4
- t4>t3>t1>t2
- t3>t1>t2>t4
- t2>t3>t1>t2
Q. The value of sin75∘+cos75∘ is
- √3√2
- 1√2
- √3−1√2
- √3+1√2
Q. If sin(θ+α)=cos(θ+α), then which of the following is/are correct?
where θ, α∈(0, π2)−{π4}
where θ, α∈(0, π2)−{π4}
- tanα=1−tanθ1+tanθ
- tanα=1+tanθ1−tanθ
- tanθ=1+tanα1−tanα
- tanθ=1−tanα1+tanα
Q. If 180∘ < θ < 270∘ and sinθ=−513 then 5 cot2θ+12tanθ+13cosecθ =
- \N
- -1
- 1
- 2
Q. If tanA=xsinB1−xcosB and tanB=ysinA1−ycosA then sinAsinB=
- xy
- yx
- x+y
- x−y
Q. If 1+sin2x1−sin2x=cot2(a+x)∀ x∈R−(nπ+π4), n∈N then the possible value of a is
- π4
- π2
- 3π4
- 3π2
Q. If sin(θ+α)=cos(θ+α), then which of the following is/are correct?
where θ, α∈(0, π2)−{π4}
where θ, α∈(0, π2)−{π4}
- tanα=1−tanθ1+tanθ
- tanα=1+tanθ1−tanθ
- tanθ=1+tanα1−tanα
- tanθ=1−tanα1+tanα
Q. The numerical value of (1−cosx)(1+cosx)(1+cot2x) is
- 0
- 1
- 2
- 3
Q. If tanA=xsinB1−xcosB and tanB=ysinA1−ycosA then sinAsinB=
- xy
- yx
- x+y
- x−y
Q. The value of sinπ5sin2π5sin3π5sin4π5 is
- 516
- 1116
- 1316
- 316
Q. If sinA=1213, cosB=−35, 0<A<π2, π<B<3π2, then the value of sin(A+B) is
- 3365
- −163
- −5665
- −6365
Q. The expression tanA1−cotA+cotA1−tanA can be written as :
- sinA⋅cosA+1
- secA⋅cosecA+1
- tanA+cotA
- secA+cosecA
Q. If sinA+sinB+sinC=0 and cosA+cosB+cosC=0, then the value of sin(A−B2) is
( where A, B, C∈[0, 2π] )
( where A, B, C∈[0, 2π] )
- 1
- 12
- 1√2
- √32
Q. If Sn=sin2πn+sin22πn+…+sin2(n−1)πn, then the value of S100 is
- 49
- 50
- 992
- 1012