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Minus one. ##sec(pi) =1/cos(pi)## ##=1/-1## ##= -1##

##sin(tan^-1(x))=x/sqrt(x^2+1)## We can use the principles of SOH-CAH-TOA: ##tan^-1(x)=theta## is the angle when ##tan(theta)=x##. Since ##tan(theta)=opposite/adjacent## we know that ##opposite=x## and ##adjacent=1##. Using we can see […]

If ##x## is a non-right angle in a right angled triangle then ##sin(x)## is the ratio of the length of the side opposite ##x## with the […]

You may see this from either the definition of ##tan## or from geometry. 1] the tangent is equal to the ratio ##sin/cos## so if you have: […]

##arctan(1/sqrt(3)) = pi/6 or 30^o## This ratio is one of the standard trigonometric triangles: ##arctan## must provide an angle in quadrants one or four; in quadrant […]

Get first the value of the of the opposite side fo get ##sin## and use the Double Angle Formula. ##costheta = 12/13## ##c^2=a^2+b^2## ##13^2=a^2 + 12^2## […]

You can consider it visually as a graph and remember the periodicity of your ##cos## meaning that it repeats itself at regular intervals (oscillating between ##+1## […]

See the proof in the explanation. ##sin^2x/(1-cosx)## Multiply the numerator and denominator both by the conjugate of the denominator. This is a fancy 1 that does […]

You will be squaring a quantity that contains both ##sinx## and ##cosx## so you will need the identity ##sin^2x+cos^2x = 1##. I would start by multiplying […]