6. Prove that tan λ cos^2 λ + sin^2λ/sin λ
= cos λ� + sin �λ
10. Prove that 1+tanθ/1-tanθ = sec^2θ+2tanθ/1-tan^2θ
17.Prove that sin^2w-cos^2w/tan w sin w + cos w tan w = cos w-cot w cos w
23. Find a counterexample to shows that the equation sec a� – cos a� = sin a� sec a� is not an identity.
PLEASE tell your classmates to use parentheses to clarify the problems!
I think you mean
(tan*cos^2 + sin^2)/sin
= (sin*cos + sin^2)/sin
= sin(cos+sin)/sin
= cos+sin
I think you mean
(1+tan)/(1-tan)
= (1+tan)^2/(1-tan^2)
= (1+2tan+tan^2)/(1-tan^2)
= (sec^2+2tan)/(1-tan^2)
I think you mean
(sin^2 - cos^2)/(tan*sin+cos*tan)
= (sin-cos)(sin+cos)/(tan(sin+cos)
= (sin-cos)/tan
= sin*cot - cos*cot
= cos-cos*cot
always try simple values for counter-examples:
a=0: 1-1 =? 0 -- yes
a=pi: -1+1 =? 0 -- yes
a=pi/4: √2 - 1/√2 =? 1/√2*√2 NO!
To prove each of these equations, we'll start by simplifying the left-hand side (LHS) of the equation and then show that it is equal to the right-hand side (RHS) of the equation. Let's solve each one step by step:
6. Prove that tan λ cos^2 λ + sin^2λ/sin λ = cos λ + sin λ:
LHS: tan λ cos^2 λ + sin^2λ/sin λ
= (sin λ/cos λ) * cos^2 λ + sin^2λ/sin λ [Using the definition of tan]
= sin λ * cos λ + sin^2λ/sin λ [Simplifying]
= sin λ * cos λ + sin λ [Canceling sin λ]
= cos λ * sin λ + sin λ [Rearranging]
= (cos λ + 1) * sin λ [Factoring]
= sin λ + cos λ [Simplifying]
Therefore, LHS = RHS and the equation is proved.
10. Prove that 1+tanθ/1-tanθ = sec^2θ+2tanθ/1-tan^2θ:
LHS: 1+tanθ/1-tanθ
= (1+tanθ)/(1-tanθ) [Simplifying]
= [(1+tanθ)/(1-tanθ)] * [(1+tanθ)/(1+tanθ)] [Multiplying by conjugate]
= (1+tanθ)^2 / (1-tan^2θ) [Using the identity (a+b)(a-b)=a^2-b^2]
= (1+tanθ)^2 / sec^2θ [Using the identity 1+tan^2θ=sec^2θ]
= (1+2tanθ+tan^2θ) / sec^2θ [Expanding]
= (sec^2θ + 2tanθ) / sec^2θ [Simplifying]
= sec^2θ + 2tanθ / 1 [Canceling sec^2θ]
Therefore, LHS = RHS and the equation is proved.
17. Prove that sin^2w-cos^2w/tan w sin w + cos w tan w = cos w-cot w cos w:
LHS: sin^2w-cos^2w/tan w sin w + cos w tan w
= (sin^2w-cos^2w) / (tan w sin w + cos w tan w) [Simplifying]
= -(cos^2w-sin^2w) / (sin w cos w + sin w/cos w) [Rearranging and flipping fractions]
= -[(cos w + sin w)(cos w - sin w)] / (sin w cos w + sin^2w/cos w) [Using difference of squares]
= -[(cos w + sin w)(cos w - sin w)] / [sin w(cos w + sin w)/cos w] [Expanding and simplifying]
= -[(cos w + sin w)(cos w - sin w)] / [cos w(cos w + sin w)] [Canceling sin w]
Now, we cancel out (cos w + sin w) from the numerator and denominator:
= -(cos w - sin w)/(cos w)
= -cos w/cos w + sin w/cos w
= -1 + tan w
Therefore, LHS = RHS and the equation is proved.
23. Find a counterexample to show that the equation sec a - cos a = sin a sec a is not an identity:
An identity holds for all values of the variable. To find a counterexample, we need to find a value of "a" for which the equation does not hold.
Let's choose a = 0 and substitute it into the equation:
LHS: sec(0) - cos(0) = 1 - 1 = 0
RHS: sin(0) * sec(0) = 0 * 1 = 0
Since LHS = RHS when a = 0, we can conclude that the equation holds true for this value of "a". Hence, it is not a counterexample.
However, if we choose a = 90 degrees, the equation becomes:
LHS: sec(90) - cos(90) = 1 - 0 = 1
RHS: sin(90) * sec(90) = 1 * infinity = infinity
Here, LHS is not equal to RHS. Therefore, a = 90 degrees is a counterexample that proves that the equation sec a - cos a = sin a sec a is not an identity.
Hope this helps! Let me know if you have any further questions.