Question

Please help me with these, oh sweet jesus

Answer 1

Answer:

77.  $\cot^{6} x = \cot^{4} x \csc^{2}x - \cot^{4} x$Proved

78.  $\sec^{4}x \tan^{2} x = \sec^{2}x [\tan^{2}x + \tan^{4}x ]$ Proved

79. $\cos^{3} x\sin^{2} x = [\sin^{2}x - \sin^{4}x] \cos x$ Proved.

80. $\sin^{4}x - \cos^{4}x = 1 - 2\cos^{2}x + 2 \cos^{4} x$ Proved.

Step-by-step explanation:

77. Left hand side

= $\cot^{6} x$

= $\cot^{4} x \times \cot^{2} x$

= $\cot^{4}x [\csc^{2}x - 1]$  

{Since we know, $\csc^{2} x - \cot^{2}x = 1$}

= $\cot^{4} x \csc^{2}x - \cot^{4} x$  

= Right hand side (Proved)

78. Left hand side

= $\sec^{4}x \tan^{2} x$

= $\sec^{2} x [1 + \tan^{2}x] \tan^{2} x$  

{Since $\sec^{2}x - \tan^{2}x = 1$}

= $\sec^{2}x [\tan^{2}x + \tan^{4}x ]$

= Right hand side (Proved)

79. Left hand side  

= $\cos^{3} x\sin^{2} x$

= $\cos x[1 - \sin^{2} x] \sin^{2} x$

{Since $\sin^{2}x + \cos^{2} x = 1$}

= $[\sin^{2}x - \sin^{4}x] \cos x$

= Right hand side

80. Left hand side  

= $\sin^{4}x - \cos^{4}x$

= $[\sin^{2}x + \cos^{2}x]^{2} - 2\sin^{2} x \cos^{2}x$

{Since $\sin^{2}x + \cos^{2} x = 1$}

= $1 - 2\cos^{2} x[1 - \cos^{2}x ]$

= $1 - 2\cos^{2}x + 2 \cos^{4} x$

= Right hand side. (Proved)